The context:

Calculate the string tension in the situation where two boxes are stacked on a slope and moving with uniform velocity, given that the kinetic friction coefficient between the heavier (bottom) box and the slope is 0.444, whereas the static friction coefficient between boxes is 0.8.

My logic of solving the given problem:

Note: By "horizontal" I mean in the direction of motion

If the static friction coefficient is large enough(which it is), the top box will not slide and the horizontal component of top boxes' gravitational force can be disregarded because static friction cancels it. Then, the tension will simply be the horizontal component of bottom box minus the friction, which takes into account the vertical component of gravity of the top box.

$$ T = F_{g,m1,x} - \mu * (F_{g,m2,y} + F_{g,m1,y})$$

My question:

As it turns out, my logic is wrong, and the horizontal component of the top boxes' gravitational force does not disappear, so the correct answer is

$$ T = F_{g,m1,x} + \textbf{F}_{g,m2,x} - \mu * (F_{g,m2,y} + F_{g,m1,y})$$

What is the argumentation that static friction does not cancel out the horizontal component of the top box?

  • $\begingroup$ Hi Bruno and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. $\endgroup$ Commented Jul 15, 2019 at 16:39
  • $\begingroup$ Looks to me that this is one of the types of HW questions we do want here: OP solved a problem incorrectly based on an assumption and wants to know why his assumption doesn't hold. $\endgroup$
    – Kyle Kanos
    Commented Jul 16, 2019 at 12:24

2 Answers 2


The missing step in your reasoning is the fact that while the static friction on the upper block (by the lower block) cancels the horizontal component of its weight, there is also static friction applied on the lower block by the upper block. Due to the third law of Newton, this would be the same as the static friction on the upper block by the lower block but in the opposite direction. So, since the static friction on the upper block is the opposite of the horizontal component the weight of the upper block, the static friction on the lower block will be precisely equal to the horizontal component of the weight of the upper block. That is how the weight of the upper block enters into the picture in your way of reasoning.

A more direct way to approach the problem would be to treat both the blocks as a single block because there is no relative motion between them. This approach is explained in the answer by Bob D.


Since the two boxes are moving as one you can consider them as one box of mass 80 kg.

Now since they are moving at constant velocity there is no net force acting on the boxes. The tension force simply equals the kinetic friction force plus the component of weight acting down parallel to the surface

From there you should be able to solve the problem (we don’t solve homework problems)

Hope this helps

  • $\begingroup$ I do understand that logic, what I fail to realize is why is it that the horizontal component of gravity of the top box has effect on the string? Does static friction not cancel it? $\endgroup$ Commented Jul 15, 2019 at 16:31
  • $\begingroup$ @BrunoRibarić You are mixing up which object you are focusing on. Just because for one object certain forces cancel out does not mean those forces become irrelevant. There is a static friction force acting on top of the bottom block, which you know must be equal to the horizontal weight component of the first block. $\endgroup$ Commented Jul 15, 2019 at 17:02
  • $\begingroup$ @Bruno Ribaric It is because in this problem the top box “sticks “ to the bottom box so it’s horizontal component becomes part of the total horizontal component $\endgroup$
    – Bob D
    Commented Jul 15, 2019 at 17:25

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