Static Equilibrium Problem from Halliday, Resnick

[mahdilm]

Hello all,

I was wondering if someone could please help with this problem. I have no idea how to do it as no lengths are provided like the practice examples that were used to demonstrate the concept in the readings. Anybody please

http://img224.imageshack.us/img224/7914/phys12pa.jpg

 

[sarujin]

If the system is in equilibrium then two statements must be true:
The sum of the torques must be zero,
and the sum of the forces must be zero.

The lengths of everything do not matter; If you choose your axis of rotation along the line connecting the two blocks then your torque equation is:
T1cos(phi)*L+50N*L-T3cos(theta)*L-40N*L=0, you can take the common factor of L, the distance between your axis of rotation and the forces, out of your equations. Then use the second part of the equilibrium equations, F(sum)=0.

 

[mahdilm]

If the system is in equilibrium then two statements must be true:
The sum of the torques must be zero,
and the sum of the forces must be zero.
The lengths of everything do not matter; If you choose your axis of rotation along the line connecting the two blocks then your torque equation is:
T1cos(phi)*L+50N*L-T3cos(theta)*L-40N*L=0, you can take the common factor of L, the distance between your axis of rotation and the forces, out of your equations. Then use the second part of the equilibrium equations, F(sum)=0.

I still don't understand

What part of the horizontal line (ie. T2) have you used as your origin for rotation. Also from what point to what point is considered 'L'?

Thanks

 

[sarujin]

If all of the forces are acting on the ends of the string T2, then they are all equidistant from the center of the string. That distance L does not matter, however, because the net torque is zero allowing you to cancel out the common factor L.

 

[mahdilm]

Ok.

I also noticed you used T1Cos(phi) for the torque equation. Is that the X component or the Y?

 

[sarujin]

The x component of the tensions provide no torque on the string because they act through our chosen axis of rotation.

Looking back now, the problem might be solved easier if you chose the axis of rotation at one of the endpoints of T2, so you can cancel one of the tensions entirely. In fact that may be the only way to solve the problem.

 

[mahdilm]

The x component of the tensions provide no torque on the string because they act through our chosen axis of rotation.
Looking back now, the problem might be solved easier if you chose the axis of rotation at one of the endpoints of T2, so you can cancel one of the tensions entirely. In fact that may be the only way to solve the problem.

Could you elaborate if possible?

 

[sarujin]

Instead of taking the axis of rotation in the midpoint of the string T2, take it at the right side. The length of the attaching string is still L. Doing this eliminates the torques -T3cos(theta)*L and 50N*L. The resulting equation is simply T1cos(phi)*L-40N*L=0. You know phi, and L cancels out once again as a common factor.

For the next parts, remember to use the second part of equilibrium: that the sum of the forces = 0.

 

[mahdilm]

Instead of taking the axis of rotation in the midpoint of the string T2, take it at the right side. The length of the attaching string is still L. Doing this eliminates the torques -T3cos(theta)*L and 50N*L. The resulting equation is simply T1cos(phi)*L-40N*L=0. You know phi, and L cancels out once again as a common factor.
For the next parts, remember to use the second part of equilibrium: that the sum of the forces = 0.

I think I partially understand now, but why did you assume the length 'L' of the string T2 was also distance from the origin to the 40N point?

 

[Pyrrhus]

If you notice T1 vector and the weight of block A vector is concurrent on their node, and T2 vector and the weight of block B vector on their node. T2 vector intersects both nodes, you can work from your left to the right.