Understanding Torque: How Balancing a Dumbbell Changes the Difficulty

[brotherbobby]

Homework Statement

A man holds a dumbbell in two different positions : (1) with his arm hanging by his side and (2) with his arm outstretched. Which of his two positions are easier and why?

Relevant Equations

Balancing of forces : $\Sigma \vec F=0$. Balancing of torques : $\Sigma \tau = 0$

The figure above shows a man holding the dumbbell in the two different positions. The force $F ( = w)$ that the man has to apply is the same in both cases. So really, the two cases should be as easy or as hard. However, in the second case, the man has to balance the torque that the dumbell would apply with his shoulders as the rotating point (pivot). Is it this which makes the second case harder?

Long as forces stay equal, is it harder to use the force to apply a torque than to apply none? I am far from convinced.

 

[haruspex]

Think about how the muscular force, which is linear, turns into a torque.
However, the vertical case is not as simple as the question seems to imply. Which muscles are used in that case?

 

[brotherbobby]

First off, don't you find it strange that the force $F$ applied by the muscle is the same in both cases? Of course using $\Sigma \vec F = 0$, it has to be since the weight of the dumbbell remains the same ($F=w$). On the other hand, if you see my diagram for the second case, the force $F$ has to balance the dumbell torque (using $\Sigma \vec \tau = 0$): $F\times \frac{l}{2} = w\times l\Rightarrow F=2w$! Something is surely wrong here.

 

[haruspex]

First off, don't you find it strange that the force $F$ applied by the muscle is the same in both cases? Of course using $\Sigma \vec F = 0$, it has to be since the weight of the dumbbell remains the same ($F=w$). On the other hand, if you see my diagram for the second case, the force $F$ has to balance the dumbell torque (using $\Sigma \vec \tau = 0$): $F\times \frac{l}{2} = w\times l\Rightarrow F=2w$! Something is surely wrong here.

Oh yes, the right hand diagram is nonsense. The forces the person applies to the arm are at the shoulder. Can you break those down?

 

[brotherbobby]

Oh yes, the right hand diagram is nonsense. The forces the person applies to the arm are at the shoulder. Can you break those down?

But if the force $F$ that the arm applies is at the shoulder, how do you explain (rotational) equlibrium taking the shoulder as the turning point? You have $\boldsymbol{wl}$ as the torque for the dumbbell (at the shoulder) and 0 for the arm force $F$ ($=w$) also at the shoulder.

 

[PeroK]

But if the force $F$ that the arm applies is at the shoulder, how do you explain (rotational) equlibrium taking the shoulder as the turning point? You have $\boldsymbol{wl}$ as the torque for the dumbbell (at the shoulder) and 0 for the arm force $F$ ($=w$) also at the shoulder.

Forget the dumbbell. It's impossible to raise your arm!

 

[haruspex]

But if the force F that the arm applies is at the shoulder

Don't restrict yourself to a single force.

 

[brotherbobby]

Forget the dumbbell. It's impossible to raise your arm!

Let's see. What if I extended my arm and applied a force equal to its weight $w$ at the shoulder? The weight of the arm would act at the centre of the arm while my force (= weight) would act my shoulder. Of course rotational equilibrium is violated. How is it then that we all can extend our arms outstretched and keep it there?

Let's take the dumbbell. Sure people can hold dumbbells too with their arms outstretched. Raise their arms too still holding it, using more force. I just can't see how both translational and rotational equilibrium can be satisfied here, unless we assume that there is a force $2F$ acting on the center of the arm going up and a force $F$ acting at the shoulder joint down ($F=w)$, along with the weight $w$ of the dumbell acting at the hand. Both equilibriua hold. But what makes this more difficult than holding the dubbell by one's side, arm hanging?

 

[haruspex]

Let's see. What if I extended my arm and applied a force equal to its weight ww at the shoulder? The weight of the arm would act at the centre of the arm while my force (= weight) would act my shoulder. Of course rotational equilibrium is violated. How is it then that we all can extend our arms outstretched and keep it there?

Let's take the dumbbell. Sure people can hold dumbbells too with their arms outstretched. Raise their arms too still holding it, using more force. I just can't see how both translational and rotational equilibrium can be satisfied here, unless we assume that there is a force 2F2F acting on the center of the arm going up and a force FF acting at the shoulder joint down (F=w)F=w), along with the weight ww of the dumbell acting at the hand. Both equilibriua hold. But what makes this more difficult than holding the dubbell by one's side, arm hanging?

Consider the muscle and bones as separate components. Where are they in relation to each other?

 

[PeroK]

Let's see. What if I extended my arm and applied a force equal to its weight $w$ at the shoulder? The weight of the arm would act at the centre of the arm while my force (= weight) would act my shoulder. Of course rotational equilibrium is violated. How is it then that we all can extend our arms outstretched and keep it there?

Let's take the dumbbell. Sure people can hold dumbbells too with their arms outstretched. Raise their arms too still holding it, using more force. I just can't see how both translational and rotational equilibrium can be satisfied here, unless we assume that there is a force $2F$ acting on the center of the arm going up and a force $F$ acting at the shoulder joint down ($F=w)$, along with the weight $w$ of the dumbell acting at the hand. Both equilibriua hold. But what makes this more difficult than holding the dubbell by one's side, arm hanging?

See @haruspex hint above.

What I would do is try to design a mechanical arm. Just a simple crane. What do you need that is missing from your analysis?