A 1,300–N uniform boom at = 67.0° to the horizontal is supported by a cable at a

Question

A 1,300–N uniform boom at = 67.0° to the horizontal is supported by a cable at an angle = 23.0° to the horizontal as shown in the figure below. The boom is pivoted at the bottom, and an object of weight w = 2,450 N hangs from its top.

(a) Find the tension in the support cable.
kN

(b) Find the components of the reaction force exerted by the pivot on the boom. (Assume the positive x-direction is to the right and the positive y-direction is upward. Include the sign of the value in your answer.)

Explanation / Answer

If the length of the beam is L then the torque applied by the 2450 N weight is
2450 L cos (67)

The torque applied by the beam will be 1300 * L /2 * cos(67) ( the value of 1/2 occurs because the centre of mass is half way along the beam)

The total torque is
2450 L cos(67) + 1300*L/2 cos(67)
= 3100 L cos (67)

This torque can only be countered by the cable. The cable is perpendicular to the boom so the torque it applies is T *3 L/4


Now 3/4 T * L = 3100 L cos (67)
T = 4/3 *3100 cos(67)
= 1615.02 N

(b)

Now you should be able to find all the components.
Horizontally the only force on the beam, and hence the pivot, is the cable, which must be opposed by the pivot.
Fx= T cos(23) = 1615.02*cos(23) = 1486.6N = 1.486kN (which must oppose the direction of the cable.)

Vertically the total weight pulls down i.e 3750 N down
and the cable pulls up with a force of Fy = T sin(23)

So the force on the pin is 3750 - T sin(23) in the downward direction

= 3750 - 1486.6*0,390 = 3169.13 N

= 3.169kN
Hence the pin pushes upwards with this exact same force ( every action has equal and opposite reaction.)

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