(a) A block of mass m = 4.90 kg is suspended as shown in the diagram below. Assu
ID: 1772595 • Letter: #
Question
(a) A block of mass m = 4.90 kg is suspended as shown in the diagram below. Assume the pulley to be frictionless and the mass of the strings to be negligible. If the system is in equilibrium, what will be the reading of the spring scale in newtons?
(b) Two blocks each of mass m = 4.90 kg are connected as shown in the diagram below. Assume the pulley to be frictionless and the mass of the strings to be negligible. If the system is in equilibrium, what will be the reading of the spring scale in newtons?
Explanation / Answer
a)
for the hanging block :
T = tension force in the string in upward direction
mg = weight in down direction
T = mg = 4.90 x 9.8 = 48.02 N
Fs = spring reading
along the horizontal direction , using equilibrium of force
Fs = T
Fs = 48.02 N
b)
Fs = T + T = 48.02 + 48.02 = 96.04 N
c)
Fs = T = mg Sin31 = 48.03 Sin31 = 24.7 N
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