Question

. A load of 250 kg is hung by a crane’s cable. The load is pulled by a horizontal force such that the cable makes a 300 angle to the vertical plane. If the load is in the equilibrium, calculate the magnitude of the tension in the cable.​

Answer 1

If the load is in the equilibrium, the magnitude of the tension in the cable is equal to 1,414.5 Newton.

Given the following data:

• Mass of load = 250 kg

• Angle of inclination = 30°

Scientific data:

• Acceleration due to gravity = 9.8 $m/s^2$

To calculate the magnitude of the tension in the cable, if the load is in the equilibrium:

First of all, we would determine the tension caused by the horizontal component of the force:

$\sum F_y ; Tcos\theta - mg=0\\\\Tcos\theta = mg\\\\T=\frac{mg}{cos\theta} \\\\T=\frac{250 \times 9.8}{cos30} \\\\T= \frac{2450}{0.8660}$

T = 2,829.1 Newton

The magnitude of the tension in the cable is given by:

$\sum F_x ; F - Tsin\theta = 0\\\\F = Tsin\theta\\\\F = 2829.1sin30\\\\F = 2829.1 \times 0.5$

F = 1,414.5 Newton.

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