- #1

- 22

- 0

## Homework Statement

A uniform horizontal beam of length 8.00 m and weight 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0 degrees with the horizontal. If a 600-N man stands 2.00 m from the wall, find the tension in the cable and the force exerted by the wall on the beam at the pivot.[/B]

## Homework Equations

∑Fx = 0

ΣFy = 0

Σtorque = 0

R = force exerted by wall at the pivot

T = tension from cable

## The Attempt at a Solution

∑Fx = R cos θ - T cos 53° = 0

ΣFy = R sin θ + T sin 53° - (weight of man) - (weight of beam) = 0

→ R sin θ + T sin 53° - 600 N - 200 N = 0

Σtorque = (weight of man) * (d1) + R(d3) - T(d2) = 0

→ (600) * (2 m) + R (4 sin θ) - T (4 sin 53°) = 0

up = positive

down = negative

→ = positive

← = negative

clockwise = positive

counterclockwise = negative

d1 = ⊥ distance from the line of action of force (weight of man) to the pivot point

d2 = ⊥ distance from the line of action of force (tension - T) to the pivot point

d3 = ⊥ distance from the line of action of force (force - R) to the pivot point

Total length of beam = 8 m so 8m/2 = 4 m

since man is 2 m from the wall, and half of 8 m is 4 m, therefore, the ⊥ distance from weight of man to center of mass (pivot) should be 2 m.

Through trig., d3 = 4 sin θ and d2 = 4 sin 53°

I am trying to solve this problem by designating the pivot point as the center of mass of the horizontal beam.

I have solved this problem by choosing the pivot at the pin connection. However, I am having trouble arriving at the same answer if I choose the pivot at the center of mass of the beam. I rearranged the equations to find the unknowns; however, I do not get the correct answer. I know that the answer should be the same regardless of where the pivot point is located. Any help would be greatly appreciated. Thanks.