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I'm very confused on this problem. A pulley has two hanging blocks on either side with one having a greater mass. To find the torque, do I have to add the tensions of both the blocks or subtract the tensions, the positive being the hanging block that generates a counterclockwise spin? I thought I needed to subtract the tensions, but for some reason I'm only getting the right answer when I add the two tensions.

The question states to find the acceleration of block A which has a mass of 4.5 kg attached to a pulley with a radius of 0.14 meters, and 0.4 kg*m^2 for moment of inertia. Then there is another block B attached to the other end with mass 2 kg.

I keep getting the wrong answer when I solve this question by subtracting tensions to get torque, but keep getting the answer right when adding tensions to get torque???

Is it that the tensions which are written in terms of mass, gravity, and acceleration have acceleration that already signify which tension is positive and negative?

What I have for tension is T(a) - M(a)*g = -M(a)*a and T(b) - M(b)*g = M(b)*a

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When you have a situation like you describe, with block A on the left and block B on the right, then the tension in A will cause a counter-clockwise torque, and the tension from B will cause a clockwise torque. Now you need to take this direction into account exactly once. So either:

  • you subtract the two tensions, and then apply a net tension with a clockwise torque, OR
  • you say that one tension applies a clockwise torque, and the other an anti-clockwise torque

The net torque in the first case is

$$\Gamma_{net}=(T_a-T_b)\cdot R$$

and in the second case:

$$\Gamma_{net} = T_a \cdot R - T_b \cdot R$$

as you can see, these are the same.

From your description, it sounds like you are doing the right thing; let me just write out how I would tackle it from here - let's see if I get a different answer than you.

Using the conventions that is A on the left, B on the right, acceleration is positive when A moves down (it is heavier), and torque is counterclockwise. Then

$$\begin{align}T_a &= m_a(g-a)\\ T_b &= m_b(g+a)\\ \Gamma &= (T_a-T_b)\cdot R \\ &= (m_a(g-a) - m_b(g+a))\cdot R\end{align}$$ $$\Gamma = ((m_a - m_b) g - (m_a + m_b) a)\cdot R\tag1$$

Now we know that this torque is what accelerates the pulley which has moment of inertia $I$:

$$\Gamma = I \dot \omega = I \frac{a}{R}\tag2$$

Combining these two equations,

$$I \frac{a}{R}=((m_a - m_b) g - (m_a + m_b) a)\cdot R\\ a\left(\frac{I}{R} + (m_a + m_b)R\right) = (m_a - m_b) g\cdot R\\ a = \frac{(m_a-m_b)gR^2}{I+(m_a+m_b)R^2}=\frac{(4.5-2.0)*9.8*0.14^2}{0.4 + (4.5+2.0)*0.14^2}=0.91\; \rm{m/s^2}$$

Is that not what you are getting?

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  • $\begingroup$ That is what I was doing but it's wrong. My question mean, does torque = tension(a) + tension(b) or tension(a)-tension(b) $\endgroup$ – Goldname Nov 4 '15 at 2:49
  • $\begingroup$ for some strange reason, I'm getting the wrong answer when I subtract, and the correct answer when I add. $\endgroup$ – Goldname Nov 4 '15 at 2:49
  • $\begingroup$ How are you calculating the tensions? Recognize that as the mass B accelerates up, the tension on B is greater than $m_b g$ while the tension in A, which is accelerating down, has to be less than $m_a g$. And the difference between these tensions is the force that gives rise to the torque that accelerates the pulley. Perhaps you need to show more of your work so we can figure out where you are going wrong. $\endgroup$ – Floris Nov 4 '15 at 2:51
  • $\begingroup$ I believe I am calculating the tensions correctly. What I have is T(a) - M(a)*g = -M(a)*a and T(b) - M(b)*g = M(b)*a $\endgroup$ – Goldname Nov 4 '15 at 2:55
  • $\begingroup$ Thank you, That's the right answer, but why are you doing T(a) - T(b)? Isn't T(b) positive? $\endgroup$ – Goldname Nov 4 '15 at 3:38

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