Equilibrium Crane: Solve M1 & θ, α for T, H, V

In summary, to solve for the tension T in the cable, the horizontal component of the force exerted on the strut by the hinge, and the vertical component of the force exerted on the strut by the hinge, the system in the figure is in equilibrium with a mass M1 of 230.0 kg hanging from the end of a uniform strut with a mass of 54.0 kg, held at an angle θ = 48.0° with respect to the horizontal and supported by a cable at an angle α = 29.7° with respect to the horizontal. By applying the equations ∑Fx = 0, ∑Fy = 0, H-T = 0, V- Ws -
  • #1
shadowice
26
0

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.

prob17a.gif

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)
visual.jpg


Homework Equations


[tex]\sum[/tex]Fx = 0
[tex]\sum[/tex]Fy = 0
H-T = 0
V- Ws - W = 0
(I called point A the bottom left forgot to label sorry)
[tex]\sum[/tex]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 I am 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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  • #2
shadowice said:

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


[tex]\sum[/tex]Fx = 0
[tex]\sum[/tex]Fy = 0
H-T = 0
V- Ws - W = 0
(I called point A the bottom left forgot to label sorry)
[tex]\sum[/tex]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 I am 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?
 
  • #3


As a scientist, my first step would be to identify all the forces acting on the system and draw a free body diagram to visualize the problem. The forces acting on the system are the weight of the hanging mass (M1), the weight of the strut (Ms), the tension in the cable (T), and the horizontal and vertical components of the force exerted by the hinge (H and V).

Next, I would use the given angles (θ and α) to break down the forces into their components. For example, the horizontal component of the tension in the cable can be found using the formula Tcosα, and the vertical component can be found using Tsinα. Similarly, the horizontal and vertical components of the weight of the strut can be found using the angle θ.

Using the equilibrium equations, \sumFx = 0 and \sumFy = 0, I would set up equations for the forces in the x and y directions. Since the system is in equilibrium, the sum of all the forces in each direction must be equal to zero.

For part A, the magnitude of the tension T can be found by setting the sum of the forces in the y direction equal to zero. This will give us an equation with two unknowns (T and V), which can be solved using basic algebra.

For parts B and C, the horizontal and vertical components of the force exerted by the hinge, H and V, can be found by setting the sum of the forces in the x and y directions equal to zero, respectively. Again, this will give us equations with two unknowns (H and V) that can be solved using algebra.

In summary, to solve for T, H, and V, we need to use the given angles to break down the forces into their components and then use the equilibrium equations to set up and solve for the unknowns.
 

1. How does an equilibrium crane work?

An equilibrium crane works by using the principles of equilibrium and torque to balance the weight of the load being lifted with the counterbalance weight. The crane arm is designed to pivot at a specific point, allowing it to remain stable and balanced even when the load is moved horizontally or vertically.

2. What is M1 and how is it calculated?

M1 refers to the moment of the load being lifted, which is the product of the load's weight and the distance from the pivot point to the center of mass of the load. It is calculated by multiplying the load's weight (in Newtons) by the distance (in meters) from the pivot point to the center of mass of the load.

3. What is theta (θ) and how does it affect equilibrium?

Theta (θ) is the angle between the crane arm and the horizontal ground. It affects equilibrium by determining the direction and magnitude of the horizontal and vertical forces acting on the crane. If theta is not balanced, the crane will not be in equilibrium and may tip over.

4. What is alpha (α) and how does it impact the crane's stability?

Alpha (α) is the angle between the crane arm and the counterbalance weight. It impacts the crane's stability by affecting the magnitude and position of the counterbalance weight, which is crucial for balancing the load and keeping the crane in equilibrium.

5. How do I solve for T, H, V in the equilibrium crane equation?

The equilibrium crane equation is T = (M1sinθ)/sin(α+θ). To solve for T, H, and V, you will need to know the values for M1, θ, and α. Plug these values into the equation and solve for T. To find H, use the equation H = Tsin(α+θ)/sinθ. To find V, use the equation V = Tcos(α+θ)-M1cosθ.

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