Two blocks of unequal mass are connected by a string over a smooth pulley. If the coefficient of kinetic friction is $\mu_k$, what angle $\theta$ of the incline allows the masses to move at a constant speed? enter image description here

So here is my attempt:

Essentially we have two 1D problems for each respective mass. So, for $v_x, v_y$ to be constant, that requires that $\ddot{x} = 0,$ and $\ddot{y} = 0$.

First, I'll consider the block of mass $m$ which moves vertically: $m \ddot{y} = T - mg = 0 \implies$ $T = mg$.

Since both masses are connected by the same string we can assume that the tensile force acting on each mass is identical.

Now, for the block of mass $2m$ which moves only horizontally:

$(2m) \ddot{x} = 2mg \sin \theta - T - F_f = 0$.

Given that $F_f := \mu F_N$, and this block does not move vertically, $F_N = 2mg \cos \theta$. This gives us that:

$2mg \sin \theta - mg - \mu_k (2mg \cos \theta) = 0$, or $\, 2\sin \theta - 1 = \mu_k (2 \cos \theta)$.

I'm not really sure how to proceed from here. The only thing I can think of doing is dividing both sides by $2 \cos \theta$, which simplifies to:

$\tan \theta - \frac{1}{2 \cos \theta} = \mu_k$, but I am not sure if this is really any better.

Any help is very greatly appreciated. Thanks!

EDIT: I just want to know if I have done the physics correctly...

  • 1
    $\begingroup$ If your solution is correct, then you've actually already solved the problem... The only thing missing is algebraic. I didn't check your calculations, but the equation you've gotten to is not easy to solve. $\endgroup$ Aug 30, 2016 at 19:22
  • $\begingroup$ Right... the issue is I am not sure if what I've done is exactly 100% correct. I'm actually a math major taking a theoretical/classical mechanics class and the last time I had intro. mechanics was 3 years ago.. So I'm quite rusty on the fundamentals. $\endgroup$
    – Javier
    Aug 30, 2016 at 19:30
  • $\begingroup$ What is your question? What 'hint' are you asking for? As QB says, you've got an equation; simplifying it further is math, not physics. $\endgroup$ Aug 30, 2016 at 19:32
  • $\begingroup$ Did I analyze the mechanics of the system correctly? $\endgroup$
    – Javier
    Aug 30, 2016 at 19:35

1 Answer 1


The equation you obtained (2sinθ−1=μk(2cosθ)) is not difficult to solve: for example, you take squares of both parts of the equation, use the formula for the sum of squares of sine and cosine and solve a quadratic equation with respect to sine. Or, alternatively, you can express the difference of the left and right parts via a sine of a sum of angles, which is a standard way to solve such equations.

However, there may be a problem with the physics of your solution. You implicitly assume that your right block moves downwards, however, for some values of the angle it may move upwards, and then you will have an opposite sign for friction. Moreover, you should consider a case where the blocks are stationary, then you will have an inequality for friction.


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