This is due by 11: 59 pm tonight.

a horizontal force f1 (of magnitude 20 n) is applied to a box that is at rest on a level surface. even with f1 applied, the box does not slide. consider the case where the magnitude of the downward force f2 is increased and complete the following using the drop-down lists:

1. the magnitude of the friction force acting on the box: (increases/decreases/remains the same)

2. the magnitude of the normal force from the tabletop acting on the box: (increases/decreases/remains the same)

3. the magnitude of the maximum possible static friction force, fs, max: (increases/decreases/remains the same)

1. increases

2. increases

3. increases

Explanation:

Part 1:

First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:

F1 - fs = 0.

And this friction force fs is:

fs = Nμs,

where μs is the static coefficient of friction, and N is the normal force.

Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:

N = mg + F2.

So, F2 is increasing, that means fs is increasing too.

Part 2:

As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.

Part 3:

In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.

a straight line sloped

report!

explanation:

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