1. Your choice of axis or rotation apparently affects the values of torque. How
ID: 2033488 • Letter: 1
Question
1. Your choice of axis or rotation apparently affects the values of torque. How certain are you that you are not getting different results with different choices of axes of rotation. Can you show mathematically moving the axis of rotation will not change net torque? From Part III, sum the weight of all of the force hangers and the meter stick. Compare this total weight with the calculated equilibrant force that you found was needed to keep the system in balance after you removed the fulcrum. Explain why these numbers are the same or different. A car pitches down in front when the brakes are applied sharply. Explain this observation in terms of torques. 2. 3.Explanation / Answer
According to Question 1.
STATEMENT:
For Instance, Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn that where the force is applied and how the force is applied is just as important as how much force is applied when we want to make something rotate. This tutorial discusses the dynamics of an object rotating about a fixed axis and introduces the concepts of torque and moment of inertia. These concepts allows us to get a better understanding of why pushing a door towards its hinges is not very a very effective way to make it open, why using a longer wrench makes it easier to loosen a tight bolt, etc.
Proof According to above case :
Consider a square with its axis of rotation in the lower left corner and with side length = L. There is no accelerating frame (the square is not on a truck). Initially it has a net torque of 0. Weight and normal force both act with the same moment arm, same magnitude, and opposite directions. (positive is counterclockwise) Net torque = 0 = (normal force * (L/2)) - (weight * (L/2)) Now consider a force being applied towards the left and acting at the midpoint of the right side. If the force being applied is small, there is zero net torque. But where does the torque countering this newly applied torque come from? It can't come from static friction since that acts through the axis of rotation. The weight acts through the same point it did before so it doesn't seem to be that. All that's left is the normal force. It must move its point of action at some distance d to the left of the midpoint of the bottom side, thereby no longer perfectly countering the weight's torque. Net torque = 0 = (force of push * (L/2)) + (normal force * ((L/2)-d)) - (weight * (L/2)) This is the only scenario that seems to account for a zero net torque: a wandering normal force. Of course as the force of the push increases, the distance "d" that the normal force acts on increases until it reaches (L/2). Then tipping occurs.
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