A weight (with a mass of 62 kg) is suspended from a point near the right-hand en

Question

A weight (with a mass of 62 kg) is suspended from a point near the right-hand end of a uniform boom with a mass of 46 kg . To support the uniform boom a cable runs from this same point to a wall (the left-hand vertical coordinate in the ?gure) and by a pivot on the same wall at an elevation of 6.5 m.

This figure is drawn to scale.

Calculate the tension T in the cable. the acceleration of gravity is 9.8 m/s^2. Answer in units of N.

A weight (with a mass of 62 kg) is suspended from a point near the right-hand end of a uniform boom with a mass of 46 kg . To support the uniform boom a cable runs from this same point to a wall (the left-hand vertical coordinate in the ?gure) and by a pivot on the same wall at an elevation of 6.5 m.

Explanation / Answer

let theta is the angle between T and horizontal rod


theta = tan^-1(3.5/6.5)

theta = 28.3 degrres


the system is in equlibrium.so, net torque is zero.



T*sin(28.3)*6.5 - 46*9.8*4.5 - 62*9.8*6.5 = 0


T = (46*9.8*4.5 + 62*9.8*6.5)/(sin(28.3)*6.5)

T = 1940 N

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