TPR Hyper Physics 133

[Doghead]

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Can anyone explain how they got the distance of 40cm, the second part of this question:

A uniform meter stick weighing 20 N has a 50-N weight on its left end and a 30-N weight on its right end. The bar is hung from a rope. What is the tension in the rope and how far from the left end of the bar should the rope be attached so that the stick remains level?

 

[DrknoSDN]

This is another balancing of torques problem.
The mass of the uniform meter stick is producing a force down at it's center of gravity (50 cm from the edges).

If you hold the left end stationary and find what torques would occur clockwise you get the meter stick (20 N at 0.5m = 10N torque) and the weight on the right end (30N at 1.0m = 30N torque)
Total CW torque of 40 Newtons if left end is held stationary.
The tension in the rope is a constant because it opposes gravity. So tension must be 100N up.

With a total clockwise torque of 40N, the only way for torques to balance is for the 100N force to be at 0.4 meters from the right side. (100N up * 0.4m = 40N up or counterclockwise)
The actual balancing would look like (100N CCW * X meters = 40N CW) and solve for X.


You could also do the problem holding the right side of the stick stationary and you would just end up with 0.6m (from the right side) required to balance the CCW torque of the stick and the 50N mass.

 

[Doghead]

@DrknoSDN - That sums it up rather nicely, thank you very much for taking the time to lay that out for me.

 

[NextStepTutor_1]

You can also calculate this answer using the center of mass equation.

Xcm = (m1*x1) + (m2*x2) + (m3*x3) / (m1 + m2 + m3)

Xcm = (2kg*50cm) + (5kg * 0cm) + (3kg * 100cm) / (10kg)

Xcm = (300)/(10) = 30 cm from the left end of the meter stick. So the rope should be placed there b/c that point acts as if all of the mass is concentrated at that point.

Also, Force Tension = Mtotal * g = (10kg) * (10m/s^2) = 100N up.

 

[Doghead]

You can also calculate this answer using the center of mass equation.

Xcm = (m1*x1) + (m2*x2) + (m3*x3) / (m1 + m2 + m3)

Xcm = (2kg*50cm) + (5kg * 0cm) + (3kg * 100cm) / (10kg)

Xcm = (300)/(10) = 30 cm from the left end of the meter stick. So the rope should be placed there b/c that point acts as if all of the mass is concentrated at that point.

Also, Force Tension = Mtotal * g = (10kg) * (10m/s^2) = 100N up.

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