An obiect with a mass of m 143 kg is suspended by a rope f and a length of 1-7.6

Question

An obiect with a mass of m 143 kg is suspended by a rope f and a length of 1-7.69 m. The end of the boom is supported by another rope which is horizontal and attached to the wall es shown in the figure. the end of a uniform boom with a mass of M 77.3 k In The boom makes an angle of 1.40×103 N -62.1 with the vertical wall. Calculate the tension in the vertical rope. ou are correct. Your receipt no. is 160-118Previous Tries Calculate the tension in the horizontal rope. (The horizontal and the vertical ropes are not connected to each other. They are both independently attached to the end of the boom.) 67312 N Incorrect. Tries 3/12 Previous Tries Submit Anewer Post Discussion

Explanation / Answer

B)

Let the pivot point be at the point where the boom is attached the wall. The weight of the boom and the tension in the vertical rope produce clockwise torque.

Torque = Force * perpendicular distance from the pivot point.

Since these two forces are vertical, the perpendicular distance from the pivot point is the horizontal component of the distance on the beam. The weight of the boom is at the center. The distance is (7.69/2) = 3.845 meters.

Horizontal component = 3.845 * cos 62.1

Weight of M = 77.3 * 9.8 = 757.54

Torque = 757.54 * 3.845 * cos 62.1= 2912.74 * cos 62.1

For the vertical rope, horizontal component = 7.69 * cos 62.1

Torque = (143*9.8) * 7.69 * cos 62.1 = 10776.77 * cos 62.1

Total clockwise torque = 2912.74 * cos 62.1 + 10776.77 * cos 62.1 = 13,689.506 * cos 62.1

The tension in the horizontal rope produces counter clockwise torque.

Vertical distance from the pivot point = 7.69 * sin 62.1

Counter clockwise torque = T * 7.69 * sin 62.1

Counter clockwise torque = Clockwise torque

T * 7.69 * sin 62.1 = 13,689.506 * cos 62.1

Divide both sides by 7.12 * sin 70.1

T = (13,689.506 * cos 62.1)/ (7.69 * sin 62.1)   

= 942.55 N

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