Center of Gravity Question: Pretty Tough Conceptually

[cheechnchong]

THANKS FOR THE HELP!!! i solved the problem successfully (correct answer too)

[PhanthomJay]

The c.g. of the arm is given. Apply the 22N force there, and sum moments (torques) about the elbow joint to solve for M, noting that the sum of the moments must be 0 for equilibrium. Mind your plus and minus signs.

[cheechnchong]

^^kinda vague...just want to know if my work is right?

[cheechnchong]

What do i plug for W3 and it's components (of distances) seems to be the misunderstanding...does my work reflect the correct answer?

[cheechnchong]

^^^Bump, i'd like to have this ASAP!!! I'm still trying to solve it, but im stuck...

[cheechnchong]

BUUUUUUUMP!!! no one helping me???

[OlderDan]

Solving for W3 i get 1464N

I think I might be miscalculating something here...like some kinda distance--i feel that the problem is right (although conceptually i am not comfortable). if someone can point out my errors, then that'll be cool! THANKS
Does W3 mean the same thing as M? I assume it does. Pick one point (the elbow looks good to me since there is an unlabled force acting on it) and calculate the torques about that point. Be careful with your signs.

[cheechnchong]

Does W3 mean the same thing as M? I assume it does. Pick one point (the elbow looks good to me since there is an unlabled force acting on it) and calculate the torques about that point. Be careful with your signs.

W3 is a force...Why do you say be careful with signs? is W3 a negative value? oh and are my measurements in the right place...can you confirm that for me?

Im checkin this thread like every few seconds lol ... so dont be surprised with quick feedback

[OlderDan]

W3 is a force...Why do you say be careful with signs? is W3 a negative value? oh and are my measurements in the right place...can you confirm that for me?
This is not about calculating center of gravity. Only one of the things you labeled as W is a weight. There are three forces acting on the forearm: its weight, which can be taken as acting at the CM, the muscle force acting very close to the elbow, and the normal force acting on the hand. Two of those forces are known. One is not. You need to use the torques associated with those forces to write an equation based on the fact that the arm is not rotating.

[cheechnchong]

This is not about calculating center of gravity. Only one of the things you labeled as W is a weight. There are three forces acting on the forearm: its weight, which can be taken as acting at the CM, the muscle force acting very close to the elbow, and the normal force acting on the hand. Two of those forces are known. One is not. You need to use the torques associated with those forces to write an equation based on the fact that the arm is not rotating.

So the equation i am using incorrect?

[Galileo]

Equation Used: X(center gravity) = W1x1 + W2x2 + W3x3 / W1 + W2 + W3

My Work:
0.15 = [(111N)(0.3m) + (.15m)(22N) + W3(0.150m + 0.025m)] / (111N + 22N + W3)

I`m afraid that's not the way at all. You used the definition of center of gravity, where the W1, W2 and W3 are the weights acting on the objects at positions x1, x2 and x3 respectively. The M and the force from the scale are not weights.

This is a static equilibrium problem. The pivot point (elbow joint) is not moving and nothing is rotating about it. That means the total torque about the elbow joint is zero.
There are three forces at work here. All applying a torque about the elbow joint: M (which you have to find), gravity and the scale (it's pushing upwards on the hand). The torque of a force about a point is r x F, where r is the distance from the pivot point to the point where the force is applied. The total torque must be zero. From this you can find M (watch the signs)

[cheechnchong]

I`m afraid that's not the way at all. You used the definition of center of gravity, where the W1, W2 and W3 are the weights acting on the objects at positions x1, x2 and x3 respectively. The M and the force from the scale are not weights.

This is a static equilibrium problem. The pivot point (elbow joint) is not moving and nothing is rotating about it. That means the total torque about the elbow joint is zero.
There are three forces at work here. All applying a torque about the elbow joint: M (which you have to find), gravity and the scale (it's pushing upwards on the hand). The torque of a force about a point is r x F, where r is the distance from the pivot point to the point where the force is applied. The total torque must be zero. From this you can find M (watch the signs)

hmmmm ok this is very clear...i appreciate this--even older dan gets to this point thanks! well, i'll scratch everything out and begin anew--i'll let you know when i have trouble! again thanks