Balancing Weight on an 8m Pole: Find the Solution

[Shah 72]

A man and a boy hold opposite ends of 8m long pole which is horizontal. A mass of 200 kg is tied to the pole. If the man is to support four times as much weight as the boy where along the pole must the weight be tied?
Can someone please help

 

[topsquark]

Hint: Make a sketch.

Put the man on the left and the boy on the right. Let the mass hang a distance d from the man. Then the mass is a distance 8 - d from the boy. Call the force the boy exerts F. Then the man exerts 4F. The pole does not rotate so the net torque on it is 0 Nm. Pick an axis, say at the man and call a clockwise rotation positive. Then the net torque is
0 = d(4F) - (8 - d)F. Now pick an axis at the mass and see if you can find F.

-Dan

 

[Shah 72]

Hint: Make a sketch.

Put the man on the left and the boy on the right. Let the mass hang a distance d from the man. Then the mass is a distance 8 - d from the boy. Call the force the boy exerts F. Then the man exerts 4F. The pole does not rotate so the net torque on it is 0 Nm. Pick an axis, say at the man and call a clockwise rotation positive. Then the net torque is
0 = d(4F) - (8 - d)F. Now pick an axis at the mass and see if you can find F.

-Dan

Thank you so so much!

 

[Kansas Boy]

You CAN'T "find F" because F cancels out of that equation.

0= d(4F)- (8- d)F. Dividing by F

0= 4d- (8- d)= 4d- 8+ d= 5d- 8

and you can easily solve for d, not F.

Don't forget that the answer must be given as a distance from a specific person and must be stated in meters.