1. Assume it is uniform and has a mass of 238kg. For the distances assume that D
ID: 1495469 • Letter: 1
Question
1.
Assume it is uniform and has a mass of 238kg. For the distances assume that D1 = 1.80m, D2 = 4.20m, D3 = 2.98m and D4 = 1.14m. For the forces assume that f3 = 4030N, f4 = 3020N and f5 = 2100N.
a. Calculate F1 for the beam shown in figure below.
b. Calculate F2.
2. A shop sign weighing 222N is supported by a uniform 140N beam of length L = 1.88m as shown in figure below.
The guy wire is connected D = 1.30m from the backboard.
a. Find the tension in the guy wire. Assume theta = 41.6o
b. Find the horizontal force exerted by the hinge on the beam.
Explanation / Answer
ANSWER
Problem 1
First, as the beam is in equilibrium, the net force and net torque acting on the beam must be equal to zero.
Apply, Fnety = 0
F1 + F2 - f3 - f4 - f5 = 0
F1 + F2 = f3 + f4 + f5 = 4030 + 3020 + 2100
F1 + F2 = 9150 N (1)
Apply, Net torque about right end = 0
f5*D4 + f4*(D4 + D3) + f3*(D4+D3+D2) + 238*9.8*(10.12/2) - F1*(D4+D3+D2+D1) = 0
2100*1.14 + 3020*(1.14+2.98) + 4030*(1.14+2.98+4.2) + 238*9.8*(10.12/2) - F1*(1.14+2.98+4.2+1.8) = 0
F1 = 60167.944/10.12
F1 = 5945.45 N ANS.
Thus, from equation (1)
F2 = 9150 N - 5945.45 N
F2= 3204.55 N ANS.
Problem 2
Part a)
Since the beam is in horizontal position, the net torque acting on the beam is zero.
Torque due to Tension + Torque due to weight + Torque due to gym board = 0
T*D*sin(theta) -(w*(L/2)) - (weight of the Gym board*L) = 0
T*1.30*sin(41.6º) - (140*(1.88/2)) - (222*1.88) = 0
Tension T = 636.03 N ANS.
Part b)
Horizontal force is,
F*cos(theta) = 636.03*cos(41.6º)
F = 475.62 N ANS.
Regards!!!
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