(a) A block of mass m = 8.90 kg is suspended as shown in the diagram below. Assu
ID: 1481108 • Letter: #
Question
(a) A block of mass m = 8.90 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?
__________N
(b) Two blocks each of mass m = 8.90 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?
____________N
(c) A block of mass m = 8.90 kg is in equilibrium on an incline plane of angle = 33.0° when connected as shown in the diagram below. Assume 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?
________N
Explanation / Answer
a) we know reading on the scale is equal to tension in the string.
As the block is in equilibrium, net force acting on it must be zero.
Apply, net force acting block = 0
T - m*g = 0
T = m*g
= 8.9*9.8
= 87.22 N <<<<<<<<<<<<<-------------Answer
b) As the blocks are in equilibrium, net force acting on them must be zero.
Let T is the tension in the string.
Apply, net force acting block = 0
T - m*g = 0
T = m*g
= 8.9*9.8
= 87.22 N <<<<<<<<<<<<<-------------Answer
c)
As the block is in equilibrium, net force acting on it must be zero.
Apply, net force acting block = 0
T - m*g*sin(33) = 0
T = m*g*sin(33)
= 8.9*9.8*sin(33)
= 47.5 N <<<<<<<<<<<<<-------------Answer
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