Exercise 11.3 A uniform rod is 1.90 m long and has mass 1.60 kg . A 2.10 kgclamp

Question

Exercise 11.3

A uniform rod is 1.90 m long and has mass 1.60 kg . A 2.10 kgclamp is attached to the rod.

Part A

How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.25 m from the left-hand end of the rod?

Express your answer with the appropriate units.

Exercise 11.3

A uniform rod is 1.90 m long and has mass 1.60 kg . A 2.10 kgclamp is attached to the rod.

Part A

How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.25 m from the left-hand end of the rod?

Express your answer with the appropriate units.

Explanation / Answer

1.6 / 1.9 = 0.842 kg / m

left side mass 1.25 X 0.842

= 1.0525 kg

now right side mass = (1.60 - 1.0525 )

= 0.5475 kg

letf side torque = ( 1.25 / 2 ) X1.0525

= 0.6578 kg / m

right side torque = 1 / 2 (1.90   - 1.25 ) x 0.5475

0.1779 kg / m

now finding the difference = ( 0.6578 - 0.1779 )

= 0.4799 kg / m

1.25 + (0.4799 / 2.1)

= 1.4785 m

1.4785 m distance from the left

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