A person holds a uniform 12.9 kg pole horizontally. The length of the pole is L

Question

A person holds a uniform 12.9 kg pole horizontally. The length of the pole is L = 4.00 m. One hand is placed at the end (point A), and the other hand is placed at point B, which is a distance x = 0.41 m from the end at point A.

What is the force of the person's hand on the pole at point A? Indicate the direction of the force with the appropriate sign.

What is the force of the person's hand on the pole at point B? Indicate the direction of the force with the appropriate sign.

A person holds a uniform 12.9 kg pole horizontally. The length of the pole is L = 4.00 m. One hand is placed at the end (point A), and the other hand is placed at point B, which is a distance x = 0.41 m from the end at point A. What is the force of the person's hand on the pole at point A? Indicate the direction of the force with the appropriate sign. What is the force of the person's hand on the pole at point B? Indicate the direction of the force with the appropriate sign.

Explanation / Answer

a) Fa*x = m*g*(L/2 - x)

Fa = m*g*(L/2 - x)/x

= 12.9*9.8*(2-0.41)/0.41

= 490.3 N

FB = Fa + m*g

= 490.3 + 12.9*9.8

= 616.72 N (upward)

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