Physics - Torque
A unifrom pole 1.2 m long weighing 60 N is carried at its ends by two men A and B. A 250 N weight is hung .3 m from man A. What is the total weight suspended from the pole: how much weight is carried by man A. By man B?
Hi... I'm not actually a teen, but I love problems like this, so I'll help you get started. (Actually, I'll probably just solve the whole thing...) This falls into a branch of physics known as "statics", meaning nothing is moving.
First, you need to consider all forces acting on the rod. There's gravity pulling down (60 N), the weight, also pulling down (250 N), and the upward force of the men holding up the rod, call these unknown forces a and b. Now, since the rod isn't falling, the total weight held by those men is 310 N, so a + b = 310.
Now, we also know that the rod isn't rotating (around it's center, 0.6 m from the end) at all either, so let's look at the torques that are acting on the rod. The force of gravity acts uniformly across the length of the rod, so the gravitation force to the left of center equals that to the right of center, canceling itself out (since it would be in opposite directions of rotation), so we can ignore the force of gravity on the rod. There are only three forces that exert a net torque. The force from the weight, and from the two men, a and b. If we assume man A is on the left, then the torque from man A would rotate the rod clockwise, while the torques from the weight and from man B would rotate the rod counter-clockwise. So ta = tw + tb.
Now, ta = 0.6 * a, tb = 0.6 * b, and tw = 0.3 * 250 = 75. So, 0.6a = 0.6b + 75. We can divide through by 0.6 to get rid of that nasty looking decimal to get a = b + 125.
So, we know that a + b = 310, and a = b + 125. At this point, the physics is done, and we just have to do the math:
Substitute, to get (b + 125) + b = 310 => 2b + 125 = 310 => 2b = 185 => b = 92.5, therefore a = 217.5.
