1. 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
2. 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
Then
, we obtain
3. 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:
4. 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:
5. 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)):
6. 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
7. 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:
8. A 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
9. 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
10. A tension of 6000 Newtons is experienced by the elevator cable of an elevator
moving upwards with an acceleration of
Solution:
From the second Newton's law we have
. What is the mass of the elevator?
Then