Questions: In the figure, the center of gravity (CG) of the pole held by the pole vaulter is 1.9 m from the left hand, and the hands are 0.68 m apart. The mass of the pole is 5.0 kg. - Part (a) Calculate the magnitude of the force, in newtons, exerted by his right hand. - Part (b) Calculate the magnitude of the force, in newtons, exerted by his left hand.

Solution Steps

We have a pole with a mass of 5.0 kg, and its center of gravity (CG) is 1.9 m from the left hand. The hands are 0.68 m apart. We need to calculate the force exerted by the left hand, given that the force exerted by the right hand is 137.1 N.

Since the pole is in equilibrium, the sum of torques about any point is zero. We choose the left hand as the pivot point. The torque due to the weight of the pole and the torque due to the force exerted by the right hand must balance each other.

The torque due to the weight of the pole is: \[ \tau_{\text{weight}} = m \cdot g \cdot d_{\text{CG}} \] where \( m = 5.0 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( d_{\text{CG}} = 1.9 \, \text{m} \).

The torque due to the right hand is: \[ \tau_{R} = F_{R} \cdot d_{R} \] where \( F_{R} = 137.1 \, \text{N} \) and \( d_{R} = 0.68 \, \text{m} \).

Set the sum of torques around the left hand to zero: \[ m \cdot g \cdot d_{\text{CG}} = F_{R} \cdot d_{R} \]

Rearrange the equation to solve for the force exerted by the left hand, \( F_{L} \): \[ F_{L} = m \cdot g - F_{R} \]

Substitute the known values: \[ F_{L} = (5.0 \, \text{kg} \cdot 9.81 \, \text{m/s}^2) - 137.1 \, \text{N} \]

Calculate: \[ F_{L} = 49.05 \, \text{N} - 137.1 \, \text{N} = -88.05 \, \text{N} \]

Since the force is negative, it indicates that the direction of the force exerted by the left hand is opposite to the assumed direction.

Final Answer