find the acceleration experienced by each of the two objects , if the coefficient of kinetic friction between the 7kg object and
the plane is 0.25
1 answer:
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Answer:
the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N
Explanation:
Given that;
Length L = 10 m
mass of beam m_b = 150 kg; weight W_beam = 150×9.8
mass of sign m = 75 kg
distance of sign hung from the beam from the wall d = 2.50 m
angle ∅ = 60°
g = 9.8 m/s²
Now,
Torque acting at one end of the beam will be;
= Tsin∅ × L - mg(d)-W × (L/2)
for equilibrium, = 0
therefore, 0 = Tsin∅ × L - mg(d)-W × (L/2)
so we substitute
Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0
Tsin(60°) × 10 - 1837.5 - 7350 = 0
Tsin(60°) × 10 - 9187.5 = 0
Tsin(60°) × 10 = 9187.5
divide both side by 10
Tsin(60°) = 918.75
T × 0.8660 = 918.75
T = 918.75 / 0.8660
T = 1060.9 N
Therefore, the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N