# Equilibrium crane

## Homework Statement

The system in the Figure is in equilibrium. A mass M1 = 230.0 kg hangs from the end of a uniform strut which is held at an angle θ = 48.0° with respect to the horizontal. The cable supporting the strut is at angle α = 29.7° with respect to the horizontal. The strut has a mass of 54.0 kg. A. Find the magnitude of the tension T in the cable.

B. Find the magnitude of the horizontal component of the force exerted on the strut by the hinge?

C. Find the magnitude of the vertical component of the force exerted on the strut by the hinge?

My visual of how i set it up (sorry is messy did in paint) ## Homework Equations

$$\sum$$Fx = 0
$$\sum$$Fy = 0
H-T = 0
V- Ws - W = 0
(I called point A the bottom left forgot to label sorry)
$$\sum$$TA = TtT -XsWs-XwW =0
cos phi = Xs/.5L
Xs = L/2 cos phi
Xw = L cos phi
Xt = L sin phi

## The Attempt at a Solution

i got all these equations but im not really sure how to apply this since we have only done vertical and horizontal struts nothing with 2 different angles.

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## Homework Statement

The system in the Figure is in equilibrium. A mass M1 = 230.0 kg hangs from the end of a uniform strut which is held at an angle θ = 48.0° with respect to the horizontal. The cable supporting the strut is at angle α = 29.7° with respect to the horizontal. The strut has a mass of 54.0 kg.

A. Find the magnitude of the tension T in the cable.

B. Find the magnitude of the horizontal component of the force exerted on the strut by the hinge?

C. Find the magnitude of the vertical component of the force exerted on the strut by the hinge?

My visual of how i set it up (sorry is messy did in paint)

## Homework Equations

$$\sum$$Fx = 0
$$\sum$$Fy = 0
H-T = 0
V- Ws - W = 0
(I called point A the bottom left forgot to label sorry)
$$\sum$$TA = TtT -XsWs-XwW =0
cos phi = Xs/.5L
Xs = L/2 cos phi
Xw = L cos phi
Xt = L sin phi

## The Attempt at a Solution

i got all these equations but im not really sure how to apply this since we have only done vertical and horizontal struts nothing with 2 different angles.
Basically you are interested in the ∑ T = 0. Once you determine the component of T that is keeping the thing up, the rest is easy.

From your drawing then, you see that you have 3 forces.

a) the weight of the strut acting through the CoM
b) the weight of the box at the end
c) the horizontal component of Tension acting to balance these.

Having different angles is no problem then really because in each case you are interested only in the moment arm that is the projection of each of these forces through either the vertical or horizontal. You've done almost all the work OK.

You applied θ (in this case cosθ) to the vertical forces and φ to the Tension (sinφ that determines the horizontal component of T to sinθ*L the height it is applied). It looks like you have all the pieces already, so just write your equation

T*sinφ*(sinθ*L) = Mstrut*g*cosθ *(L/2) + M1*g*cosθ*L

Where L is the length of the strut, that you are not given ... but which simply cancels out right?