Uday M. answered • 09/27/19

Nuclear Engineer: B.S. UC Berkeley, M.S UC Davis, LLNL employee

First, I would recommend drawing a free body diagram and accounting for all the forces. You'd want to start with gravity (G) which of course acts straight down and the normal force (N) of the ground pushing back up on the box; both of these act in the y-direction. There is the force of the person pulling on the box which we can term an applied force (A), and there is the friction pulling backwards on the box; both of these act in the x-direction.

Remember there are two types of friction: static friction which is what stops the box from moving in the first place (SF), or kinetic friction which slows the box after it is already moving (KF). When we discuss friction, if the object starts off from rest, we need to figure out whether the applied force is enough to overcome SF. If it isn't, then the box would not move. If it is, the box will move and be hindered by KF.

Recall the equation for friction. For both SF and KF, it is μ*N, where μ is the coefficient of friction and N is the normal force. The μ value depends on the material: some materials are more 'rough' than others and cause more friction (e.g. sandpaper vs ice). The μ for SF and KF are also different: μ(SF) is higher than μ(KF) because it is harder to get an object to move from rest than it is to keep it moving.

So in order to get the friction, we first have to find N since it is in the equation. Let's write Newton's second law in the y-direction.

G + N = m * a(y)

The weight of the object is the gravity that it is experiencing. Remember that you can solve for the object's mass from the gravity. The object is not accelerating in the y-direction. Here you can solve for N. Make sure your signs are appropriate.

Now we have to figure out which is greater, SF or A.

A > SF ?

Here you can plug in the equation for SF (and then subsequently plug in the values for μ and N that you just solved for) and A which was given. Whether this inequality holds will tell you if the box begins to accelerate.