Question

1. A 12.0 kg sign is to be hung from the end of a uniform horizontal beam of length 2.50 m and mass 55.0 kg.
A vertical wire supports the beam near the end where the sign is located, and a pin attaches the beam
to a wall on the opposite end of the beam.
(a) If the pin can withstand a maximum force of 100 N, find the minimum distance from the wall that
the vertical wire can be attached to the beam.
(b) Where can the wire be attached to the beam (as measured from the wall) in order for the force in
the pin to be zero?

Answer 1

a)
Since the arrangement is in static equilibrium, the net force and the net torque about any point is zero.
Consider the vertical forces,
Np - Wb - Ws + T = 0
Np - The force on the pin, Np = 100 N
Wb - The weight of the beam, Wb = 55 * 9.81 = 539.55 N
Ws - Weight of the sign, Ws = 12 * 9.81 = 117.72 N
T - The tension in the wire.

T = Wb + Ws - Np
= 539.55 + 117.72 - 100
= 557.27 N

Consider the net torque about the pin,
Wb * L/2 + Ws * L - T * x = 0
Where L is the length of the beam and x is the distance to the vertical wire from the pin.
539.55 * 2.5/2 + 117.72 * 2.5 - 557.27 * x = 0
557.27 * x = 968.7375
x = 1.74 m

b)
Consider the net torque about the vertical wire. Consider that the vertical wire is at a distance d from the wall.
Np * d + Wb * (d - L/2) - Ws * (L - d) = 0
Given that Np = 0
Wb * (d - L/2) - Ws * (L - d) = 0
Wb * (d - L/2) = Ws * (L - d)
539.55 * (d - 1.25) = 117.72 * (2.5 - d)
539.55 * d - 674.4375 = 294.3 - 117.72 * d
657.27 * d = 968.7375
d = 1.47 m