A tightrope walker is walking between two buildings holding a pole with length L = 16.0 m, and mass m p = 20.0 kg. The daredevil grips the pole with each hand a distance d = 0.585 m from the center of the pole. A bird of mass m b = 585 g lands on the very end of the left‑hand side of the pole. Assuming the daredevil applies upward forces with the left and right hands in a direction perpendicular to the pole, what magnitude of force F left and F right must the left and right hand exert to counteract the torque of the bird?

When the tightrope walker, the pole and the bird are at equilibrium, the

sum of the force, and moments about a point are zero.

The magnitude of force exerted by the left hand is 140.21 N

The magnitude of force exerted by the right hand is 61.73 N

Reasons:

Given parameter are;

Length of the pole, L  = 16.0 m

Mass of the pole, M = 20.0 kg

Distance of the left and right hand from the center, d = 0.585 m

Mass of bird, m = 585 g

Location bird lands = End of left hand side of the pole

Required:

Force exerted by the left and right hand, , and , respectively.

Solution;

Weight of the pole, W = 20.0 kg × 9.81 m/s² = 196.2 N

Taking moment about the center of the pole gives;

Clockwise moments   =       Anticlockwise moments

        0.585·                =  0.585×9.81 × 8 + 0.585·

0.585· - 0.585·  =  0.585×9.81 × 8 = 45.9108

0.585· - 0.585·  = 45.9108...(1)

Sum of forces ∑F = 0, therefore;

↑ +↑ + ↓ + ↓ = 0

Therefore;

+ = +

Which gives;

+ = 196.2 + 0.585×9.81 ≈ 201.94

+ ≈ 201.94...(2)

Dividing equation (1) by 0.585 gives;

-  = 45.9108/0.585 = 78.48...(3)

Adding equation (1) to equation (3) gives;

2· = 201.94 + 78.48 = 280.42

The force to be exerted by the left hand, = 140.21 N

= 201.94 - = 201.94 - 140.21 = 61.73

The force to be exerted by the right hand, = 61.73 N

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F1 = 140.2[N]; F2 = 61.72[N]

Explanation:

To solve this problem, we must make a free body diagram, with the different forces and distances that apply on the Daredevil and its bar.

In the attached image is a free body diagram, with the different forces and distances that are on the equilibrium bar.

First, we performed a sum of forces at y-axis equal to zero, in this way we deduced one of the equations for the forces of the arms F1 & F2. It takes another equation, to find each of the forces.

Then we do a sum of moments equal to zero, at the point of force F1, in this way we can find an equation for Force F2. With the force F2 we replace in the first equation and we can find the force F1.

In this way the forces are:

F1 = 140.2[N]

F2 = 61.72[N]

In the attached image we can see the equations developed to find the forces F1 & F2