Prediction With the spring scale, you\'ll be measuring how much force is needed
ID: 1950051 • Letter: P
Question
Prediction With the spring scale, you'll be measuring how much force is needed to support one end of the meter stick, as the hanging weight is moved closer and closer to the spring scale. Keep in mind that the force applied to the meter stick by the string holding the hanging weight doesn't change: no matter where you place the hanging weight, in equilibrium its string will exert a force on the stick that's equal in magnitude to Wh. What do you predict will happen to the spring scale reading F, as the hanging weight is moved in from the far end of the meter stick: will it increase steadily, decrease steadily. increase and then decrease, decrease and then increase, or remain constant? Explain your reasoning in terms of torque and equilibrium.Explanation / Answer
The reading will steadily increase as the weight is moved to the left. Torque = (distance from axis) x Force in this case, the axis is the far left end of the stick, and the force of the weight is constant (as is the distance of the force reader from the axis). For the system to be in equilibrium, the torques must be equal to 0. Thus, as the weight moves to the right, it's distance from the axis is increasing, so it's torque is increasing. The spring scale must then increase the amount of force it needs to put on the meter stick.