A 1250 N uniform boom is supported by a cable perpendicular to the boom, as seen

Question

A 1250 N uniform boom is supported by a cable perpendicular to the boom, as seen in the figure below. The boom is hinged at the bottom, and an m = 2030 N weight hangs from its top. Assume the angles to be = 68.9o and = 90.0o - . Find the tension in the supporting cable. Tries 0/8 Find the x-components of the reaction force exerted on the boom by the hinge (choose to the right as positive). Tries 0/8 Find the y-components of the reaction force exerted on the boom by the hinge (choose upwards as positive).

Explanation / Answer

A 1300 N uniform boom is supported by a cable perpendicular to the boom, as in the figure below. The boom is hinged at the bottom, and a 2100 N weight hangs from its top.

Find the tension in the supporting cable. N


Find the components of the reaction force exerted on the boom by the hinge.
to the right N
upward N

here is the image
http://www.webassign.net/sercp/p8-26alt.gif (image)

You told me my previous answer was WRONG. Given that you had left out the diagram I naturally assumed that the support was at the top of the beam.

Your diagram shows a different point of connection.

All the answers are identical except that the torque applied by the cable is not T * L
It is 3/4 T * L

Now redo the answers accordingly.

If the length of the beam is L then the torque applied by the 2100 N weight is
2100 L cos (65)

The torque applied by the beam will be 1300 * L /2 * cos(65) ( the value of 1/2 occurs because the centre of mass is half way along the beam)

The total torque is
2100 L cos(65) + 1300*L/2 cos(65)
= 2750 L cos (65)

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 = 2750 L cos (65)
T = 4/3 *2750 cos(65)
= 1549 N

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(25) which must oppose the direction of the cable. ( = 1404 N)

Vertically the total weight pulls down i.e 3400 N down
and the cable pulls up with a force of Fy = T sin(25)

So the force on the pin is 3400 - T sin(25) in the downward direction

(= 3400 - 655 = 2745 N)
Hence the pin pushes upwards with this exact same force ( every action has equal and opposite reaction.)

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