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$\bullet$ In Figure $10.44,$ forces $\vec{A}, \vec{B}$ $\vec{C},$ and $\vec{D},$ each have magnitude50 $\mathrm{N}$ and act at the same point on the object. (a) What torque (magnitude and direction) does each of these forces exert on the object about point $P ?$ (b) What is the total torque about point $P ?$

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Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

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So here we're going to say counterclockwise is positive. And then we can say that the torque is equal to the force times the distance from the center of Mass. And then we can see that the torque subnet is going to be equal to positive. 50 Nunes times 0.20 sign of 60 degrees and this is equaling 8.7 Newton meters again. This is going to be a counter clockwise rotation because we chose counterclockwise to be positive for the torque at. For Target B would be equal to zero Newton per meter on DH. Then the torque it see is going to be equal. Tio Ah, negative 50 nunes Times point to sign of 60 degrees and again, Ah rather sign of 30. Decrease my apologies, and this will be negative. Five Newton meters again. This would produce a clockwise rotation, and we have the torque at D. This is going to be equal to negative 50 Newton's times 0.2, and this is giving us negative 10 Newton meters again. This is clockwise, so if we want to find the sum of the torques, will simply take the sum of a B, C and D. We know that B is going to be equal to zero, so we simply have 8.7, minus five minus 10 and this is giving us negative 6.3 new meters again. This is producing a clockwise rotation because of its negative. It'll be opposite to the direction that we chose is positive. We chose counterclockwise to be positive. So again, if it's negative, it's producing a clockwise rotation. This will be our final answer. That is the end of the solution. Thank you for watching.

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