Consider a 80 kg man and his 20 kg son balancing on a seesaw, which is a uniform board 2 m long and pivoted exactly at its center. If the boy sits at the end of the seesaw, how far from the center on the other side should the father sit? [Type the ANSWER below in proper STEPS]

When tightening a bolt, you push perpendicularly on a wrench with a force of 150 N at a distance of 0.125 m from the center of the bolt. Part (a) How much torque are you exerting on the wrench, in newton meters, relative to the center of the bolt? Part (b) What is this torque, in pound-foot?

Alunch tray is being held in one hand, as the figure illustrates. The mass of the tray itself is 0.237 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.19-kg plate of food and a 0.241-kg cup of coffee. Obtain (a) the force 7 exerted by the thumb and (b) the force exerted by the four fingers. Both forces act perpendicular to the tray, which is being held parallel to the ground. 0.400 m 0.380 m 0.240 m 0.100 m 0.0600 m henc henc

For two existing torques, what third force at a given distance from the pivot will balance them? Imagine a meter stick set up as in the figure. It hangs from a central bracket, and two hanging masses can hang from it from each of their brackets. At a third location, a force probe can either pull up or pull down on the stick, depending on what is needed to balance the stick. The mass of the meter stick is 120 g. sketch the situation (drawing r1, r2, r3, F1, F2, and F3) and determine the magnitude (value) and direction (+ or -) of each torque. Don't include the mass of a bracket that would hold the hanging mass in place; assume the mass listed is the entire mass hanging at that point. For each trial, use the principle of equilibrium (where the sum of torques is zero) to calculate the third, unknown force acting at x3