A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. Calculate the magn
itude of the tension in the wire if the angle between the cable and the horizontal is θ = 47°.
1 answer:
Answer:
T = 2010 N
Explanation:
m = mass of the uniform beam = 150 kg
Force of gravity acting on the beam at its center is given as
W = mg
W = 150 x 9.8
W = 1470 N
T = Tension force in the wire
θ = angle made by the wire with the horizontal = 47° deg
L = length of the beam
From the figure,
AC = L
BC = L/2
From the figure, using equilibrium of torque about point C
T (AC) Sin47 = W (BC)
T L Sin47 = W (L/2)
T Sin47 = W/2
T Sin47 = 1470
T = 2010 N
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Answer:
The value of tension on the cable T = 1065.6 N
Explanation:
Mass = 888 kg
Initial velocity ( u )= 0.8
Final velocity ( V ) = 0
Distance traveled before come to rest = 0.2667 m
Now use third law of motion =
- 2 a s
Put all the values in above formula we get,
⇒ 0 = - 2 × a ×0.2667
⇒ a = 1.2
This is the deceleration of the box.
Tension in the cable is given by T = F = m × a
Put all the values in above formula we get,
T = 888 × 1.2
T = 1065.6 N
This is the value of tension on the cable.