Consider the following system:

enter image description here

Where we have:

$M_a$ = Mass of A

$M_b$ = Mass of B

$R$ = Radius of the Pulley

$I$ = Moment of Inertia of the Pulley

$\mu$ = Coefficient of friction between A and table

Now, using Newton Second Law, calculate the acceleration of the block B.

I was able to calculate it using conservation of energy. But when it comes to applying net force, and resulting torque, I'm kinda lost.

Here is what I've done:

I've considered that the weight block B applies a force in one extreme of the string, and the Friction force in block A applies another force in the other extreme of the string, hence: $$ F_b = M_b g\\ F_a = - \mu M_a g $$ Then, I thought that the string is just transferring the forces to the pulley (correct me if I'm wrong to assume that), where these forces is resulting in some net torque: $$ T_{F_b} = M_b g r\\ T_{F_a} = -\mu M_a g r\\ T_{net} = gr(M_b - \mu M_a) $$ But we know, by Newton second law for rotational kinematics, that: $$ T_{net} = I \alpha $$ Then: $$ I \alpha = gr(M_b - \mu M_a)\\ \alpha = \frac{gr(M_b - \mu M_a)}{I} $$

But what I've found is angular acceleration $\alpha$. Is it going to be the same as the acceleration of block B?

Can someone please check if what I've done is indeed correct and if it's not, correct and explain to me my mistakes? Thanks!

  • 1
    $\begingroup$ I assume that where you say "spring" you actually mean "string". If the string has an elasticity that needs to be taken into account things would be a whole lot more complicated...? $\endgroup$
    – Floris
    Jan 30, 2017 at 13:52
  • $\begingroup$ Your reasoning seems to make sense. Angular acceleration is related to tangential acceleration by $\alpha \times r = a$ $\endgroup$
    – JMac
    Jan 30, 2017 at 13:52
  • 2
    $\begingroup$ @Floris Yeah, you are correct. Sorry, I'll edit it. But in that case we are not considering any elasticity to the string. $\endgroup$
    – Bruno Reis
    Jan 30, 2017 at 13:54
  • $\begingroup$ @JMac Yeah, forget about that relation. So if I multiply $r$ to my $\alpha$ I'll get the acceleration of the block B, right? $\endgroup$
    – Bruno Reis
    Jan 30, 2017 at 13:56

1 Answer 1


You are on the right path. You want to assume an acceleration $a$ (to be determined), then write the forces in terms of that.

This means that the tension between A and the pulley now is sufficient to accelerate $A$ and overcome the friction,

$$T_A = m_a a + \mu m_A g$$

The tension between B and the pulley is the force of gravity minus the force needed to accelerate B:

$$T_B = mg - ma$$

and the difference between these forces is just enough to accelerate the pulley.

Note that angular acceleration $\alpha$ is related to linear acceleration $a$ with $a = \alpha R$ where $R$ is the radius of the pulley.

Can you do it now?

  • $\begingroup$ Thank you for answering. What exactly is a tension? Can't I just think of it as a string who is transferring forces to some other object, in our case, to the pulley? $\endgroup$
    – Bruno Reis
    Jan 30, 2017 at 14:01
  • $\begingroup$ Tension is a measure of force in the string. When the string is massless, tension is the same at every point in the string that doesn't "touch" something with mass (like A, B and the pulley). At those points, the string can transfer force to the object, and the tension can change. $\endgroup$
    – Floris
    Jan 30, 2017 at 14:08
  • 1
    $\begingroup$ hmmm, I think I'm getting it. Thank you my friend!!! $\endgroup$
    – Bruno Reis
    Jan 30, 2017 at 14:19

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