13 April 2010

Dynamics

Problem 1.

If an object weighs 30 N on Earth, how much would it weigh on the moon?

Solution: 

The free fall acceleration on Earth is
The free fall acceleration on Moon is
 
The weight of the object on Earth is . From this expression we can find the mass of the object:

Then we can find the weight of the object on Moon:
 

Problem 2.

A child throws a ball downward from a tall building. Note that the ball is thrown, not dropped and disregard air resistance. What is the acceleration of the ball immediately after it leaves the child's hand?

Solution: 

After the ball leaves the child's hand it is in the air.
There is only one force acting on the ball in the air (if we disregard air resistance).
This force is the gravitational force.
This force will provide acceleration – free fall acceleration.

Therefore the acceleration of the ball is the free fall acceleration:

 
Problem 3.

You are driving along an empty straight road at a constant speed u. At some point you notice a tall wall at a distance D in front of you. Would it require a larger force to (a) continue moving straight and decelerate to a full stop before the wall, or (b) turn left or right to avoid the wall? (to make the calculation easier assume that the turn is done at a constant speed along a circular path).

Solution: 

(a) If the car continue moving straight then the acceleration should satisfy the following equation:
where - final velocity, which is 0, since the car should stop just before the wall. - initial velocity, which is equal to u. Then
If the mass of the car is m, then the force required to stop the car is

 
(b) Now the car is turning left or right to avoid the wall. This is the uniform circular motion with a radius D. The acceleration of this motion is
Then the force required to turn the car is
We can see that in the case (a) the force is 2 times smaller 

Problem 4.

How fast should the earth spin in order for a 150 lb human not to be able to walk on the ground?

Solution: 

The condition that the human cannot walk on the ground is that there is no contact between human and the ground. It means that the normal force is 0.
Let us assume that the earth is rotating with angular velocity . Then the human on the ground will have centripetal acceleration with the magnitude
Where is the radius of the earth.
This acceleration is provided by the gravitation force and the normal force, so that
Then
Then condition that the normal force is 0 (or negative) is
Then
and
 

Problem 5.
 
In the film 2001: A Space Odyssey, a wheel like space station achieves artificial gravity by spinning around its axis. If the station had a size of 2 km, how fast should it be spinning for the people inside to feel the same gravitational acceleration as on earth? 

Solution: 

If the space station is rotating with angular velocity then the acceleration (centripetal acceleration) at distance R=2 km =2000 m is given by the equation

This acceleration should be equal to the free fall acceleration. Then

and
 


 A boy of mass 40 kg wishes to play on pivoted seesaw with his dog of mass 15 kg. When the dog sits at 3 m from the pivot, where must the boy sit if the 6.5 me long board is to be balanced horizontally? 

Solution: 

We have equilibrium. It means that the net torque should be equal to 0. If - is the distance from the boy to the pivot, then the equilibrium condition becomes:



 

Problem 7.

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are: partice A = 400 kg, particle B = 500 kg, and particle C = 100 kg. Find the magnitude and direction of the net gravitational force acting on each of the three particles (the direction to the right is positive).


Solution: 

Particle A: The net gravitational force is the sum of two forces: due to particle B and due to particle C. Both forces have positive direction:



 
Particle B: The net gravitational force is the sum of two forces: due to particle A (this force is negative, which means that its direction is to the left) and due to particle C (this force is positive). Then



Particle C: The net gravitational force is the sum of two forces: due to particle A and due to particle B. Both forces are negative, then



Problem 8.

As part a of the drawing shows, two blocks are connected by a rope that passes over a set of pulleys. The block 1 has a weight of 400 N, and the block 2 has a weight of 600 N. The rope and the pulleys are massless and there is no friction.
(a) What is the acceleration of the lighter block?
(b) Suppose that the heavier block is removed, and a downward force of 600 N is provided by someone pulling on the rope, as part b of the drawing shows. Find the acceleration of the remaining block.

Solution: 

(a) The acceleration of the lighter block is equal by magnitude to the acceleration of the other block. If the tension of the rope is T, then the equations of motion for block 1 and block 2 are the following:

Then
and


 
(b) In this case the tension of the rope is given: T = 800N. Then the second Newton's law for block 1 becomes:



Problem 9.

Part a of the drawing shows a block suspended from the pulley; the tension in the rope is 80 N. Part b shows the same block being pulled up at a constant velocity. What is the tension in the rope in part b ?





Solution: 

In part (a) the equilibrium condition can be written as:

In part (b) we have the motion with constant velocity. It means that the acceleration is 0 and the net force is 0, which can be written as
From these two equations we can find
 


Problem 10. 

A uniform, solid metal disk of mass 6.0 kg and diameter 2.0 cm hangs in a horizontal plane, supported at its center by a vertical metal wire.  You find that it requires a horizontal force of 4 N tangent to the rim of the disk to turn it by 5 degrees, thus twisting the wire.  You now remove this force and release the disk from rest.
(a) What is the torsion constant for the metal wire? 
(b) Write the equation of motion for twist angle of the disk.



Solution: 

(a) The torque is related to the torsion constant (k) by the equation:

We know that when we apply the force 4 N the system is in equilibrium with a twist angle . The torque in this case is . Then

(b) To write equation of motion we need to find the moment of inertia of the disk. It has the following expression:
Then
or


Problem 11. 

A box is sliding up an incline that makes an angle of 20 degrees with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.2. The initial speed of the box at the bottom of the incline is 2 m/s. How far does the box travel along the incline before coming to rest?



Solution: 

The first part in the problem is to find an acceleration of the motion. The acceleration is due to gravitation force and the friction force and has the following form:

The second part is to write down the kinematic equations of motion. In this problem we need to use the relation between the traveled distance and initial and final (the final velocity is 0) velocities:
where s is the traveled distance. Then
 


Problem 12. 

A block weighing 80 N rests on a plane inclined at 30 degrees to the horizontal. The coefficients of static and kinetic friction are 0.2 and 0.1 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping?



Solution: 

The minimum force corresponds to the condition that the static friction force has the maximum value, which is 0.2*normal force. To find the normal force and the external force we need to write down the condition of equilibrium: the net force is 0. Then we rewrite this equation in terms of x and y-components (x axis is parallel to the plane).

The x-component of the second Newton's law has the form:

The y-component:

Then since , we obtain

Then

Problem 13. 

The tallest spot on Earth is Mt. Everest, which is 8857 m above sea level. If the radius of the Earth to sea level is 6369 km, how much does the magnitude of g change between sea level and the top of Mt. Everest?



Solution: 

The value of the free fall acceleration is
Then
Or


Problem 14. 

The value of g at the surface of the earth is 9.78 N/kg, and on the surface of Venus the magnitude of g is 8.6 N/kg. A cosmonaut has a mass of 60 kg on the surface of the earth. What will her weight be on the surface of Venus?



Solution: 

The weight is the product of the mass of cosmonaut and the free fall acceleration.
On the surface of Venus:

 

Problem 15. 

A car (m=2000 kg) is parked on a road that rises 20 degrees above the horizontal. What are the magnitudes of (1) the normal force and (2) the static frictional force that the ground exerts on the tires?



Solution: 

We need to write down the second Newton's law: the net force is 0. This is the vector equation. In the present problem there are three forces acting on the car: gravitational force, normal force, and the static friction force.

Then we rewrite the second Newton's law in terms of x and y-components (x axis is parallel to the road).

The x-component of the second Newton's law has the form:
The y-component:

From these equations we can find the normal force:
and the static friction force:
 

Problem 16. 

A rocket of mass kg is in flight. Its thrust is directed at an angle of 60 degrees above the horizontal and has a magnitude of . Find the magnitude and direction of the rockets acceleration. Give the direction as an angle above the horizontal.



Solution: 

There are two forces acting on the rocket: the first one is the gravitation force and the second one is the trust force. The net force (the sum of these two forces) will provide an acceleration of the rocket. It is easier to find the x- and y-components of the net force. If axis x is horizontal axis and axis y is a vertical axis then

Then acceleration can be found from the second Newton's law:

The magnitude of the acceleration is

The direction is characterized by the angle:

Problem 17. 

The speed of a bobsled is increasing because it has an acceleration of . At a given instant in time, the forces resisting the motion, including kinetic friction and air resistance, total 500 N. The mass of the bobsled and its riders is 300 kg.
(a) What is the magnitude of the force propelling the bobsled forward?
(b) What is the magnitude of the net force that acts on the bobsled?


Solution: 

The net force in the sum of two forces – friction (and air resistance) force and the force propelling the bobsled forward. There are also normal force and gravitational force, but they cancel each other. Then

The net force provides an acceleration of the bobsled. Then

Then we can find the force :

Problem 18. 

A person in a kayak starts paddling, and it accelerates from 0 to 0.8 miles/hour in a distance of 0.8 km. If the combined mass of the person and the kayak is 80 kg, what is the magnitude of the net force acting on the kayak?



Solution: 

At first we need to find an acceleration of the kayak (with the person). To do this we need to use the relation:

Where - the final velocity; - initial velocity; – traveled distance.

Then

 
Then from the second Newton's law we can find the magnitude of the net force:
 


Problem 19. 

A dancer is standing on one leg on a drawbridge that is about to open. Before the drawbridge starts to open, it is perfectly level with the ground. The dancer is standing still on one leg. What is the x component (horizontal component) of the friction force?



Solution: 

Since there is no motion of the dancer and there is no possible motion of the dance, then the x-component (horizontal component) of the friction force is 0. 

Problem 20. 

A block of mass 5 kg lies on a horizontal table. The block is at rest. The only forces acting on the block are the force due to gravity and the normal force from the table. What is the magnitude of the friction force?



Solution: 

Since there is no motion of the block and there is no possible motion of the block, then the friction force is 0.

Problem 21. 

Starting from rest, a skier slides 200 m down a 35 degrees slope. How much longer does the run take if the coefficient of kinetic friction is 0.3 instead of 0?



Solution: 

The acceleration of the skier on a slope is
With zero friction we have:
With friction 0.3 we have:
 
To find the traveled time we need to use the equation:
Then
 
and
 

Problem 22. 

A soccer ball of diameter 35 cm rolls without slipping at a linear speed of 2 m/s. Though how many revolutions has the soccer ball turned as it moves a linear distance of 20 m?



Solution: 

When the ball makes one turn it travels the distance

Then if the ball travels L=20 m = 2000 cm then it makes
 


Problem 23. 

A bicycle is moving at 10 m/s.  What is the angular speed of its tires if their radius is 30 cm?



Solution: 

The linear speed is related to the angular speed by the equation:

Where , . Then

 

Problem 24. 

The gravitational force that the sun exerts on the moon is perpendicular to the force that the earth exerts on the moon.
The masses are: Mass of Sun , Mass of Earth: ,, Mass of Moon: . The distance between the sun and moon is , and between the moon and the earth is .
Determine the magnitude of the net gravitational force on the moon.

Solution: 

The magnitude of the gravitation force that the sun exerts on the moon is

The magnitude of the gravitation force that the earth exerts on the moon is

Since the directions of the forces are orthogonal then the magnitude of the net force is
 

Problem 25. 

A tension of 6000 Newtons is experienced by the elevator cable of an elevator moving upwards with an acceleration of . What is the mass of the elevator?


Solution: 

From the second Newton's law we have
Then
 


Problem 26. 

What conditions are necessary for an object to move with constant velocity?
What conditions are necessary for an object to move with constant acceleration?


Solution: 

From the second Newton's law we know that:

1. The condition that the object moves with constant velocity (zero acceleration) is that the net force is 0.

2. The condition that the object moves with constant acceleration (and the acceleration is a vector, so it means that we have the same direction and the same magnitude of acceleration) is that the net force is constant – the direction and the magnitude of the net force are the same duration the motion of the object. 

Problem 27. 

Consider a wet roadway banked, where there is a coefficient of static friction of 0.40 and a coefficient of kinematic friction of 0.2 between the tires and the roadway. The radius of the curve is R=80m
(a) if the banking angle is 30,what is the maximum speed the automobile can have before sliding up the banking?
(b) What is the minimum speed the automobile can have before sliding down the banking?

Solution: 

(a) The direction of the forces and acceleration are shown in the figure. We write down the second Newton's law in terms of components:
 
X (horizontal axis)- components:
y (vertical axis)- components:

We also have the relation between and :

Then we have:
and

Then from the first equation (x-component):
Then
 
 
(b) In this case we have opposite direction of the friction force.


Then x and y components of the second Newton's law become:
 
We have the same relation between and :
Then
and
Then from the first equation (x-component):
Then


Problem 28. 

Is it easier to move a heavy box that is sitting on the ground by (a) pulling the box from a rope that makes an angle with the surface or (b) by pushing the box with a force that makes the same angle (but pointing downwards) with the surface?

Solution: 

In both cases the horizontal components of the tension force in the rope will be the same, but the vertical components will have opposite direction.


Then the normal force in the case (a) will be less than the normal force in case (b).


Then the friction force (which is proportional to the normal force) will be less in case (a) then in case (b).


Then it is easier to pull the box (case (a)) then to push it (case (b)).


Problem 29. 

The space shuttle is orbiting the Earth at a distance of about 300 km from its surface. At that distance, the gravitational acceleration is almost the same as that on the surface. How long does it take for the shuttle to complete one orbit around the Earth? Assume that the orbit is circular.


Solution: 

The gravitation force will provide the acceleration to the shuttle, which is equal

where

Then and
 

Then the period is



Problem 30. 

A spaceship is on a straight-line path between Earth and moon. At what distance from the Earth is net gravitational force on the spaceship zero?


Solution: 

Let us introduce the mass of the Earth as , the mass of the moon as , and the distance between the Earth and the moon as .

If the spaceship is at distance x from the Earth, then the gravitation force on the spaceship due to Earth is

The gravitation force on the spaceship due to moon is

Since the net gravitation force should be zero, then we have the equation:
From this equation we have:
and
 


Problem 31. 

A circus clown weighs 900 N. The coefficient of static friction between the clown's feet and the ground is 0.4. He pulls vertically downward on a rope that passes around three pulleys and is tied around his feet. What is the minimum pulling force that the clown must exert to yank his feet out from under himself?

Solution: 

There are the following forces acting on the clown:
Tension force (applied to the hand) – direction upward
Gravitational force – direction downward,
Normal force – direction upward;
Friction force – direction horizontal;
Tension force (applied to clown's legs) – direction horizontal.
Then the minimum tension force correspond to the condition that the friction force is equal to 0.4*normal force.
In terms of x (horizontal) and y (vertical) components we have:


Then





Problem 32. 

The athlete, who has a mass of 100 kg, can throw a 500 g ball with a speed of 10 m/s. The distance through which his hand moves as he accelerates the ball forward from the rest until he releases it is 1.0 m. What constant force must the athlete exert on the ball to throw it with this speed?

Solution: 

If we know the acceleration then we can find the force. Therefore the first step should be to find the acceleration of the ball. We need to use the following equation of motion with constant acceleration
The initial velocity is 0, the final velocity is 10 m/s and the travelled distance is 1 m. Then we can find acceleration
Then the force is

where the mass of the ball is 500 g = 0.5 kg. 

Problem 33. 

When a falling object is at a distance above the Earth's surface of 4 times the Earth's radius, what is its free-fall acceleration due to the Earth's gravitational force exerted on it?

Solution: 

The gravitation force has the following expression
Here is the mass of the Earth, is the radius of the Earth, is the distance above the Earth. Since then
Then acceleration is
Since then

Problem 34. 

A 10.0 kg block is towed up an inclined at with respect to the horizontal. The rope is parallel to the incline and has a tension of 100 N. Assume that the block starts from rest at the bottom of the hill, and neglect friction. How fast is the block going after moving 40 m up the hill?

Solution: 

To find the speed of the block we need to find the acceleration of the block. The speed is related to acceleration and traveled distance by the following equation (see kinematics)

where . The acceleration is given by the equation

Then
 

Problem 35. 

A 0.01 kg object is moving in a plane. The x and y coordinates of the object are given by and y . Find the net force acting on the object at t=2 s.

Solution: 

To find the net force we need to find acceleration of the object. The acceleration is the second derivative of coordinate of the object. Then the x and y components of acceleration are given by the following expressions
Then at t=2 s we have
Then the magnitude of acceleration is
Then the net force is

Problem 36. 

An automobile weighing 3200 lbs. is on a road which rises 10 ft. for each 100 feet of road. What force tends to move the car down the hill?

Solution: 

All variables in the problem should be expressed in the same units (the same system of units). It is better to use the SI units. In the SI system of units, the mass should be measured in kg and the distance is in meters. Then:
The force acting on the car are the following: gravitation force, normal force, friction force, and the force pulling the force up along the hill. The friction force is the rolling friction force, which is small. The normal force does not have a component along the hill (the normal force is orthogonal to the incline). Then there is only one force pulling the car down the hill. This is the gravitational force, shown in the picture below. The magnitude of the gravitation force is
 

Then from the picture of the incline we can find the component of the gravitational force along the inline. This component can be expressed in terms of the angle of incline, :
To find the angle of incline we need to use the right triangle, shown in the picture above:
Then


Problem 37. 

At 20 meters long rope attached at the top and the bottom of a flag pole is pulled 2 meters away from the pole by a 100 newton force acting at right angles to the pole at its mid point. What is the tension on the segments of the rope on each side of the 100 newton force?

Solution: 

In problem we need to use the equilibrium condition for point A: the vector sum of all forces is zero. There are three forces acting on point A: an external 100 N force and two tension forces . The directions of the tension forces are along the ropes. These forces are shown in the figure below.



The y component of the external force is zero and the y-components of the tension forces cancel each other. Then the condition that the x-component of the sum of all these forces is zero takes the form:
Then
The angle can be found from the right triangle (with lengths of 10 m and 2 m), shown in the figure:

Then


Problem 38.

A crane cable that is capable of withstanding 22,000 N is attached by a hook to  a 2,000 kg block that is resting on the ground. The cable initially starts lifting the block at the maximum acceleration that the cable can withstand for 4 sec.  It then continues to raise the block at constant velocity for further 2 sec. At this time the block slips off the hook at the end of the cable.
Calculate:
(1) the tension in the cable when the block is moving at constant velocity;
(2) the maximum acceleration that the cable can withstand;
(3) the maximum height that the block reaches above ground.

Solution: 

(1) When the block is moving at constant velocity the acceleration of the block is 0. Then from the second Newton's law we get that the net force on the block is 0:
There are two forces acting on the block: gravitation force, pointing upward, and the tension in the cable. The forces have opposite directions, then the magnitude of the forces should be the same (the vector sum of these two forces is zero):
Therefore the tension in the cable is 19600 N.

(2) The maximum acceleration should be found from the condition that the tension in the cable has its maximum value (22000 N). There are two forces acting on the block: gravitation force and the tension in the cable. Then the second Newton's law takes the form:
Then we rewrite this vector equation in terms of y-components (see figure below):
Then

 

(3) In the part we need to use the kinematics equations. First we have motion with constant acceleration (we know its value from part (2)). The equations, which describe this motion, are the following:
The initial velocity is zero: , the initial height is zero: . Then we have
The block travels . Then at this moment
 
Then the block moves with constant velocity. This velocity is . The block travels for 2 seconds. The corresponding distance is
 
Then the final height of the block is
 

Problem 39.

A body of mass 10g is set to rotate in a circular path by means of a string 200 cm long. If it makes 3 complete revolutions in 2s, find the tension of the string.

Solution: 

The first step is to convert all variables into SI units:
mass should be measures in kg: ;
length should be measured in meters: .
 
Now we can solve the problem. The body moves in a circle with constant speed. This is the motion with acceleration – centripetal acceleration. The acceleration is pointing towards the center of the circle. The magnitude of acceleration is
Here is the speed of the body (the speed is constant) and is the radius of the circle, which is equal to the length of the string. Therefore, to find the magnitude of the centripetal acceleration we need to know the speed of the body.
We know that the body makes 3 complete revolutions in 2 seconds. During one revolution the body travels a distance of , then during 3 revolutions it travels a distance of in . Then the speed of the body is
Now we can find the magnitude of centripetal acceleration:
 
The centripetal acceleration is provided by the tension in the string. Then from the second Newton's law we obtain the tension in the string:
 

Problem 40.

A particle moves in a circle of radius 1 m. Its linear speed is given by , where t is in second and v in meter/second. Find the radial and tangential acceleration at

Solution: 

The tangential acceleration is defined as the change of the speed (magnitude of the velocity) of the particle. Therefore the tangential acceleration is


The radial acceleration is centripetal acceleration, which changes the direction of velocity. The centripetal acceleration is determined by the speed (not the change of the speed) and the radius of circular orbit:

The velocity at is , then
 

Problem 41. 

A 1.5 kg mass is attached to the end of a 90 cm string. The system is whirled in a horizontal circular path. The maximum tension that the string can withstand is 400 N. What is the maximum number of revolutions per minute allowed if the string is not to break?

Solution: 

In this problem we need to use the second Newton's law and the fact that this is a motion with acceleration – centripetal acceleration. There are two forces acting on the mass: gravitational force and the tension force. The vector sum of these two forces provides the acceleration of the mass.
 
..................................................(1)
The centripetal acceleration is pointing towards the center of circle (rotation in a circle).
To characterize the direction of the string we introduce angle .

 
Now we can rewrite equation (1) in terms of x and y components:
x-component:
..................................................(2)
 
y-component:
 
Now we need to use the expression for centripetal acceleration in terms of the angular velocity and the radius of circular orbit:
.......................................................(3)
The radius of circular orbit can be expressed in terms of the length of the string and the angle :
We substitute this expression in equations (2) and (3) and obtain
We can divide both sides of this equation by and obtain

We know the tension has the maximum value: 400 N. Then we can find the angular velocity:
 
If we know the angular velocity then we can find the period of rotation:

This is the time of one revolution. To find the number of revolutions per one minute (60 seconds) we need to divide 60 seconds by the time of one revolution:
 

Problem 42. 

The force required to stretch Hooke's Law spring varies from 0 N to 65 N as we stretch the spring by moving one end 6.3 cm from its unstressed position. Find the force constant of the spring. Answer in units of N/m.

Solution: 

We understand from the formulation of the problem that the spring force is 65 N when the spring is stretched by , which in SI units is .

At the same time from the Hooke's law we know that the spring force is
where is a force constant (force constant) of the spring. We substitute the known values on the above expression and obtain
Then


Problem 43. 

A 100 kg load is placed on a 40 kg stretcher to be carried by two persons. The stretcher is 2m long and the load is placed 0.6m from the left end. How much force must each person exert to carry the stretcher?

Solution: 

In this problem we need to use the condition of equilibrium for stretcher, which mean there is no translational motion and there is no rotation of the stretcher.
First, we need to show all forces acting on stretcher. It is important to show not only the directions of the forces but also their application points. We characterize the position of the application point in terms of the distance to point A (shown in the figure below):

1. Force due to person 1. Application point of the force is one end of the stretcher (see figure). The distance to point A is .

2. Force due to person 2. Application point of the force is another end of the stretcher (see figure). The application point is point A.

3. Gravitational force on the stretcher. Application point of the force is the center of the mass of the stretcher, which is exactly in the middle of the stretcher – point O. Since the length of the stretcher is 2 m then the distance between point O and point A is .

4. Gravitational force on the load. Application point of the force is the center of the mass of the load. The distance between the load and the point A is .

 
The stretcher is in equilibrium (static or dynamics equilibrium). We have two equations, which describe the condition of equilibrium:
(1) no translational motion – the net force is zero:
We rewrite this vector equation in terms of y-components (axis y is shown in the figure):
Then
............................................(1)
 

(2) no rotation – the net torque is zero. We can choose any point to write down the condition that the net torque about this point is zero. We choose point A. Then the net torque about point A is
We substitute the known values and obtain
Now we can find force :

And then force (from equation (1)):
 

Problem 44. 

A painter weighing 900 N stands on a massless plank 5 m long that is supported at each end by a stepladder. If he stands 1m from one end of the plank, what force is exerted by each stepladder?

Solution: 

This problem is similar to Problem 43. The main difference is that now the plank is massless. Therefore, in the present problem we have only three forces acting on the plank.
We need to use the condition of equilibrium for the plank: no translational motion and no rotation of the palnk.
First, we need to show all forces acting on the plank. It is important to show not only the directions of the forces but also their application points. We characterize the position of the application point in terms of the distance to point A (shown in the figure below):

1. Force due to stepladder 1. Application point of the force is one end of the plank (see figure). The distance to point A is .

2. Force due to stepladder 2. Application point of the force is another end of the plank (see figure) – this is the point A.

3. Gravitational force on the painter. Application point of the force is the center of the mass of the painter. The distance between the painter and the point A is .

 
The plank is in equilibrium (static or dynamics equilibrium). There are two conditions of equilibrium:

(1) no translational motion – the net force is zero:
We rewrite this vector equation in terms of y-components (axis y is shown in the figure):
Then
..............................................................(1)
 
(2) no rotation – the net torque is zero. We can choose any point to write down the condition that the net torque about this point is zero. We choose point A. Then the net torque about point A is zero:
We substitute the known values and obtain
Now we can find force :

And then force (from equation (1)) is


Problem 45. 

A 20 kg child and a 30 kg child sit at opposite ends of a 4 m seesaw that is pivoted at its center. Where should another 20 kg child sit in order to balance the seesaw?

Solution: 

In this problem we need to use only one equilibrium condition: no rotation of the seesaw (the seesaw is balanced). The second equilibrium condition (no translational motion) needs to be used only if we need to find the normal force at the pivot.
We can introduce any point as the point of possible rotation. We consider point O – the center of the seesaw – as the point of possible rotation. There is no rotation about point O, which means that the net torque about point O is zero.
First, we need to show all forces acting on the seesaw:

1. Gravitational force (more exactly it is normal force, which is equal to the gravitational force) on the 20 kg child. Application point of the force is one end of the seesaw (see figure). The distance to point O is .

2. Gravitational force (more exactly it is normal force, which is equal to the gravitational force) on the 30 kg child. Application point of the force is the other end of the seesaw (see figure). The distance to point O is .

3. We place another 20 kg child on the seesaw. We characterize the position of the child by the distance to point O (this is an unknown distance): . The corresponding force is the gravi tational force on the child.
 
 
There is no rotation about point O. Then the net torque about point O is zero:

We substitute the known values and obtain

From this equation we can find the position of the 20 kg child:
 

Fluids and elasticity

Problem 1.

A cylindrical vessel of radius 0.1 meter is filled with water to a height of 0.5 meter. It has a capillary tube 0.15 meter long and 0.0002 meter radius fixed horizontally at its bottom. Find the time in which the water level will fall to a height of 0.2 meter.
Problem 2. 
Calculate the amount of energy needed to break a drop of water of diameter 0.001 meter into 1000000 droplets of equal sizes. The surface tension of water is 0.0072 N/m. 
Problem 3. 
A force F is applied on a square plate of side L. If the percentage error in determination of L is 3% and that in F is 4%. what is the permissible error in pressure? 
Problem 4. 
(Physics for scientists and engineers - Serway and Jewett - Chapter 12 - Problem 31)
A walkway suspended across a hotel lobby is supported at numerous points along its edges by a vertical cable above each point and a vertical column underneath. The steel cable is 1.27 cm in diameter and is 5.75 m long before loading. The aluminum column is a hollow cylinder with an inside diameter of 16.14 cm, an outside diameter of 16.24 cm, and unloaded length of 3.25 m. When the walkway exerts a load force of 8500 N on one of the support points, how much does the point move down?