For sure it must be true that T1+T2=W or the stick would either move up or down.

Question

For sure it must be true that T1+T2=W or the stick would either move up or down. But also the torques that the forces exert must must cancel out as well. Let us consider rotations about point X at the left end. Force T1 does not cause any tendency for the stick to rotate about this point so it has no torque about X. On the other hand, the force W causes the stick to rotate clockwise about X and the force T2 causes it to rotate counterclockwise about X.

For equilibrium the clockwise rotation or torque must equal the counterclockwise. If this were not true the object would start spinning.

The torque i fW about X is just W*Lw and the torque of T2 about X us T2*L2.

So for rotational equlibrium W*Lw=T2*L2

The distances Lw & L2 are called the lever arms of the forces.

1) Using the preceding picture assume that the stick is a meter stick and that T1=100N and T2=150N. Find how large W must be for the stick to be in translational equilibrium and where it must be applied along the stick for the stick to be in rotaional equilibrium. Be sure to start working the problem by drawing a FBD showing the forces on the stick.

2) Now assume the stick has a mass of 0.1kg and repeat the above exercise. Note that the force of gravity is effectively acting at the CM of the stick, in this case the middle or 50cm mark.

PLEASE SHOW ALL STEPS

For sure it must be true that T1+T2=W or the stick would either move up or down. But also the torques that the forces exert must must cancel out as well. Let us consider rotations about point X at the left end. Force T1 does not cause any tendency for the stick to rotate about this point so it has no torque about X. On the other hand, the force W causes the stick to rotate clockwise about X and the force T2 causes it to rotate counterclockwise about X. For equilibrium the clockwise rotation or torque must equal the counterclockwise. If this were not true the object would start spinning. The torque i fW about X is just W*Lw and the torque of T2 about X us T2*L2. So for rotational equlibrium W*Lw=T2*L2 The distances Lw & L2 are called the lever arms of the forces. 1) Using the preceding picture assume that the stick is a meter stick and that T1=100N and T2=150N. Find how large W must be for the stick to be in translational equilibrium and where it must be applied along the stick for the stick to be in rotaional equilibrium. Be sure to start working the problem by drawing a FBD showing the forces on the stick. 2) Now assume the stick has a mass of 0.1kg and repeat the above exercise. Note that the force of gravity is effectively acting at the CM of the stick, in this case the middle or 50cm mark.

Explanation / Answer

1)W=T1+T2=250N ; W*x=T2*L ; x=0.6 m ; 2) W=T1+T2-mg = 249.019 m ; W*x + mg*L/2 = T2*L ; x=0.600 m

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