Part A
Suppose a 10 kg cubic box is at rest on a horizontal surface. There is friction between the box and the surface characterized by a coefficient of static friction equal to 0.50. If one person pushes on the box with a force F1 of 25 N directed due north and a second person pushes on the box with a force F2 = 25 N directed due south, what is the magnitude and direction of the force of static friction acting on the box?
System: Box as point particle subject to external influences from the earth (gravity) the surface (normal force and friction) and the two people (applied forces).
Model: Point Particle Dynamics.
Approach: To determine the force of static friction, we first find the net force in the absence of friction. We first draw the situation. A top view (physical representation) and a side view (free body diagram) of the box ignoring any contribution from friction are shown here.
The net force parallel to the surface in the absence of friction is then:
\begin
[ \sum_{F \ne F_{f}} F_
= 0 ]
[\sum_{F \ne F_{f}} F_
= F_
- F_
= 0 ]\end
Thus, the net force along the surface is zero without the influence of static friction, and so the static friction force will also be 0.
Part B
Suppose a 10 kg cubic box is at rest on a horizontal surface. There is friction between the box and the surface characterized by a coefficient of static friction equal to 0.50. If one person pushes on the box with a force F1 of 25 N directed due north and a second person pushes on the box with a force F2 = 25 N directed due east, what is the magnitude and direction of the force of static friction acting on the box?
System: Box as point particle subject to external influences from the earth (gravity) the surface (normal force and friction) and the two people (applied forces).
Model: Point Particle Dynamics.
Approach: To determine the force of static friction, we first find the net force in the absence of friction. We first draw the situation. A top view (physical representation) and a side view (free body diagram) of the box ignoring any contribution from friction are shown here.
The net force parallel to the surface in the absence of friction is then:
\begin
[ \sum_{F \ne F_{f}} F_
= F_
= \mbox
]
[\sum_{F \ne F_{f}} F_
= F_
= \mbox
]\end
In order to prevent the box from moving, then, static friction would have to satisfy:
\begin
[ F_
= - \mbox
\hat
- \mbox
\hat
= \mbox
^
\mbox
.]\end
We're not finished yet!
We must now check that this needed friction force is compatible with the static friction limit. Newton's 2nd Law for the z direction tells us:
\begin
[ \sum F_
= N - mg = ma_
]\end
Note that friction from an xy surface cannot act in the z direction.
We know that the box will remain on the surface, so az = 0. Thus,
\begin
[ N = mg = \mbox
]\end
With this information, we can evaluate the limit:
\begin
[ F_
\le \mu_
N = \mbox
]\end
Since 35.4 N < 49 N, we conclude that the friction force is indeed 35.4 N at 45° S of W.