Torques and Tensions

TA1068

Homework Statement

A horizontal uniform bar of mass m and length L is hung horizontally on two vertical strings. String 1 is attached to the end of the bar and string 2 is attached a distance L/4 from the other end. A monkey of mass m/2 walks from one end of the bar to the other. Find the tension T_1 in string 1 at the moment that the monkey is halfway between the ends of the bar.

Homework Equations

$$\tau = rFsin\theta$$
$$F_thrust = ma_t = mr\alpha$$

The Attempt at a Solution

In all honesty, I'm really not too sure where to start.
I drew a diagram of what is going on. I know at least that the tensions are going to add up to (3/2)mg, since there is the m of the bar pulling down and also the m/2 of the monkey pulling down and the system is in equilibrium.

So: $$T_1 + T_2 = 1.5mg$$
Becomes: $$T_1 = 1.5mg - T_2$$

Do I have to choose a pivot point? I was thinking where $$T_2$$ connects to the bar, because it was an answer to a question leading up to this, but I'm not exactly sure why? What effect does placing the pivot point here have in relation to $$T_2$$?

Any help would be greatly appreciated!

TA1068
Oops! I forgot to attach my image. Here it is attached, and if that doesn't work I uploaded it as well.

img229.imageshack.us/img229/1795/tensionwm2.jpg

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Homework Helper
Hi TA1068,

Whereever you place your pivot point, if there is a force (or more than one) acting at that point then that force produces no torque about that point (because the lever arm is zero). So that force or forces will not appear in the torque equation for that pivot point.

Because of that, quite often a good choice for the pivot point is at the place where an unknown force is acting because your equation will then have one less unknown variable.

So choosing the pivot point at the place where T2 pulls on the bar is a good choice. What do you get?

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