Supporting a trapezoid between two wedges

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SUMMARY

The discussion focuses on analyzing the forces acting on a trapezoid (B) positioned between two wedges. The normal forces are defined as Nr = 5 sin 45° and Nl = 5 sin 60°, while the frictional forces are Fr = 5 sin 45° and Fl = 5 sin 30°. Participants emphasize the importance of drawing accurate force diagrams to visualize the equilibrium of forces acting on both the trapezoid and the wedge (A). The conversation highlights the necessity of vectorial addition to understand the balance of forces effectively.

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Homework Statement
Find the weight W that can be supported by the applied force of 5 kN. Mu = .25 at all surfaces
Relevant Equations
F = mu N, where mu = coefficient of static friction, N = normal force
Let Nr = normal force on the right side of the trapezoid B and Fr = the force of friction on the right side of B.
Let Nl = normal force on the left side of B and Fr = the frictional force on the left surface of B

so Nr = 5 sin45 and Fr = 5 sin45 =
and Nl = 5 sin 60, Fl = 5 sin 30.
 

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All surfaces would include the underside of A.
Consider balance of forces on A and balance of forces on B.
 
Yes, thank you. But I am having trouble drawing a force diagram which makes sense at the two surfaces of B.
 
Tom Hammer said:
Yes, thank you. But I am having trouble drawing a force diagram which makes sense at the two surfaces of B.
 
Tom Hammer said:
Yes, thank you. But I am having trouble drawing a force diagram which makes sense at the two surfaces of B.
Not sure why. What forces act on B?
What forces act on A?
If the system slips, can it do so without sliding at all contact surfaces?
 
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Tom Hammer said:
Yes, thank you. But I am having trouble drawing a force diagram which makes sense at the two surfaces of B.
Why do you find it difficult?
Forgetting for a moment about block A, only three forces acting perpendicularly to each of the top and side surfaces of B are needed to keep in equilibrium.
Try a vectorial addition of those forces, which should form a triangle.
 

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