Question

In the diagram above a mass of 160.9 kg is suspended by a cable held away from a wall by a solid strut. The angle in the diagram is 25.3 degrees. Answer the questions that follow.

Answer 1

Hi there!

To solve, we can begin by doing a summation of torques. We can place the fulcrum at the point at which the strut and wall make contact to make solving easier.

Recall that:

`\tau = r * F` (Cross Product)

τ = Torque (Nm)

r = distance from lever arm (m)

F = Force (N)

This can also be written as:

`\tau = rFsin\theta`

For a system to be in rotational equilibrium:

`\tau _(cw) = \tau_(ccw)`

In this instance, we only have the torques of the force of gravity and the vertical component of the cable's tension that are in opposite directions. Additionally, these torques are PERPENDICULAR to the strut, so sin(θ) = 1.

Let 'r' represent the length of the bar.

`rTsin\theta = rF_g\\\\Tsin\theta = F_g\\\\T = (F_g)/(sin\theta) = (9.8 * 160.9)/(sin(25.3)) = \boxed{3689.695 N}`

To find the compression force exerted by the wall on the strut, we can do a summation of forces in the horizontal direction.

Let F(W) represent the force on the strut by the wall, and Tx represent the horizontal component of the tension. These two sum up to zero since the system is in equilibrium.

`\Sigma F_x = T_x - F_W`

`0 = T_x - F_W\\\\F_W = T_x\\\\F_W = Tcos\theta = 3689.695cos(25.3) = \boxed{3335.789 N}`

**This is assuming the strut has no mass.