Matter & Interactions 2nd ed. Practice Problems
Aaron Titus | High Point University
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(N)=# of solutions
1230002     Force on a box that is pushed in equilbirum across a floor.     1230002
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Question

 

You push a 15-kg box across a somewhat smooth concrete floor as shown below.


Figure: A box being pushed across the floor.

You push with a force of 20 N at an angle, with respect to the horizontal, of . If the box moves in the +x direction with a constant speed, what is the force by the floor on the box? (Note: the gravitational force of Earth on the box is .)

 



Solution

 

(a) The box has a constant speed in a constant direction. This means that its momentum is constant.

(b) Its motion can be used to analyze the forces on the box by applying the momentum principle. Here are the steps:

(1) define the system:

(2) apply the momentum principle to the system. The box's momentum is constant. Therefore, $\Delta \vec{p}=0$.

Therefore, the net force on the box is zero. Note that even though the box is moving its momentum is NOT changing. That's why the net force on the box is zero. Just because it's moving does NOT mean that there's a net force on the object. If the net force is not zero, then the box will speed up, slow down, and/or change direction. In this case, the box is doing none of those.

To calculate an individual force acting on the system, you must apply the definition of net force as the vector sum of all forces acting on the system. Here are the steps:

(1) List the objects in the surroundings that exert forces on the system. This list includes any objects that make contact with the system and any objects that exert "force at a distance" such as a gravitational force or an electrostatic force, for example. Here are the objects in the surroundings that exert forces on the system.

(2) Sketch a free-body diagram showing the forces acting on the system with approximate magnitudes and directions if they are known. In this case, the only significant forces on the box are the gravitational force of Earth on the box, me pushing the box, and the force by the floor due to contact with the box. A side view of a force diagram for the girl is shown below.


Figure: The force diagram showing the forces on the box, from the side view.

Note that the force by the floor on the box must be in the direction shown so that when adding the vectors tail to head, the resultant (or net force) is zero. Verify that the forces in the force diagram sum to give the correct net force.

Another way to sketch it is to first draw the force by Earth and the force by me tail to head. The net force on the box is zero, so the force by the floor, drawn tail to head, must be drawn from the head of the second vector to the tail of the first vector.


Figure: The net force on the box is zero. Drawing the forces tail to head, they must add to zero.

(3) Apply the definition of net force and solve for the unknown force. Sometimes you have to do this in component form and sometimes you can do it in vector form. We'll solve it in vector form.

First, we need to express as a vector. We are given its magnitude and its direction. Sketch the vector along with the +x and +y axes. Find the angle it makes with the +x and +y axes, and use direction cosines.


Figure: Angles needed to express the force by me on the box as a vector.

The angle that the vector makes with the +x axis is , and the angle is . The reason that both angles are negative is that they are both measured clockwise from the +x axis and +y axis, respectively.

and

Thus, which points to the right and downward, in agreement with our picture. Using the definition of the net force, we can solve for the force by the floor on the box.

The floor exerts a small force in the -x direction and a large force in the +y direction. This is in agreement with our picture which shows that the force by the floor on the box points to the left and upward, with a larger y-component than x-component.


Figure: The force by the floor on the box.

If we apply the definition of the net force in component form, you can more easily make some important observations. For example, consider the y-component of the net force.

The above equation says that the upward component of the force by the floor on the box is equal to the negative sum of the gravitational force on the box and the downward component of the force by me on the box. In other words, the harder I push down on the box as I'm moving it, the harder the floor has to push upward to keep the net force zero in the y-direction. If I do not push downward on the box at all, then the floor pushes upward with the same magnitude force as the gravitational force on the box. However, when I push the box downward, the floor balances that by pushing upward with a greater force then gravitation alone. The free-body diagram below shows only the y-components of the forces. You can see that the upward force by the floor on the box balances the sum of the gravitational force and the downward force by me on the box.


Figure: Free-body diagram showing only the y-components of the forces.