Question

The small crane is mounted along the side of a pickup bed and facilitates the handling of heavy loads. When the boom elevation angle is θ=40°the force in the hydraulic cylinder BC is 4.5 kN, and this force applied at point C is in the direction from B to C (the cylinder is in compression). Determine the moment of this 4.5kN force about the boom

Answer 1

this is the solution of your question.

Answer 2

Answer:

To determine the moment of the 4.5 kN force about the boom, we need to calculate the torque (also known as the moment) generated by the force. The torque is equal to the force multiplied by the perpendicular distance from the axis of rotation (the boom) to the line of action of the force.

Let's assume that point B is located at the origin (0, 0), and the boom is along the positive x-axis. Then, point C can be represented as (d, h), where d is the distance from B to C along the x-axis, and h is the distance from B to C along the y-axis.

Using trigonometry, we can find d and h as follows:

d = d

h = d * tan(40°)

The moment about the boom is equal to the force multiplied by the perpendicular distance from the axis of rotation to the line of action of the force, which is h:

Moment = Force * h = 4.5 kN * d * tan(40°)

Thus, the moment of the 4.5 kN force about the boom is equal to 4.5 kN * d * tan(40°).

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