# Static Rigid Bodies Mech Question

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#1
I'm a bit stuck on this question! Any help would be great!

A beam AB has weight 40 N and length 3 m. The beam is freely hinged at the end A to a vertical wall. The beam is held in equilibrium at an angle of 60° to the wall by a rope.

One end of the rope is attached to the point C on the beam, where AC = 2 m. The other end of the rope is attached to a point D on the wall, where D is vertically above A.

The rope is perpendicular to the beam, as shown in Figure 1. The rope and the beam lie in a vertical plane that is perpendicular to the wall. The beam is modelled as a uniform rod and the rope as a light inextensible string.

Using the model, find

(a) the tension in the rope,

(b) the magnitude of the resultant force acting on the beam at A.

If the rope was not modelled as being light,

(c) state how this would affect the tension along the rope, explaining your answer.

Last edited by chefvicious; 1 month ago
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1 month ago
#2
(Original post by chefvicious)
I'm a bit stuck on this question! Any help would be great!

A beam AB has weight 40 N and length 3 m. The beam is freely hinged at the end A to a vertical wall. The beam is held in equilibrium at an angle of 60° to the wall by a rope.

One end of the rope is attached to the point C on the beam, where AC = 2 m. The other end of the rope is attached to a point D on the wall, where D is vertically above A.

The rope is perpendicular to the beam, as shown in Figure 1. The rope and the beam lie in a vertical plane that is perpendicular to the wall. The beam is modelled as a uniform rod and the rope as a light inextensible string.

Using the model, find

(a) the tension in the rope,

(b) the magnitude of the resultant force acting on the beam at A.

If the rope was not modelled as being light,

(c) state how this would affect the tension along the rope, explaining your answer.

What have you tried, where are you stuck, ...
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#3
(Original post by mqb2766)
What have you tried, where are you stuck, ...
for part a :

t = 40cos30 = 20sqrt(3)

I'm not sure if that is correct but I'm stuck on part b
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1 month ago
#4
(Original post by chefvicious)
for part a :

t = 40cos30 = 20sqrt(3)

I'm not sure if that is correct but I'm stuck on part b
The usual way to approach questions is to put all the forces on there then take moments, resolve horizontally and vertically, ... Maybe upload your diagram and describe how you got part a) then think about part b).
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1 month ago
#5
If anyone has an account on Chegg the answer is on there
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1 month ago
#6
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1 month ago
#7
Not helpful. I assumed that the plan was to try to improve the OP's understanding of mechanics.
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1 month ago
#8
(Original post by chefvicious)
for part a :

t = 40cos30 = 20sqrt(3)

I'm not sure if that is correct but I'm stuck on part b
To get that you appear to have resolved perpendicular to the rod. That means that you should have included the force on the rod at A.

To avoid that, take moments about A and then the force at A won't come into the equation.
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#9
(Original post by tiny hobbit)
To get that you appear to have resolved perpendicular to the rod. That means that you should have included the force on the rod at A.

To avoid that, take moments about A and then the force at A won't come into the equation.
thank you for the help!!
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