- #1
goldfish9776
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Member warned about proper use of the template and showing an attempt at solution.
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Homework Help Overview
The discussion revolves around the forces acting on a block on an inclined plane, specifically focusing on the total downward force and the role of tension in a rope connected to the block. Participants are analyzing the equilibrium conditions of the block and the implications of various forces, including gravitational, tension, and frictional forces.
Discussion Character
- Mixed
Approaches and Questions Raised
- Participants are questioning whether to consider the tension in the rope and how it affects the equilibrium of the block. There are discussions about calculating the total downward forces and the implications of kinetic friction being mentioned in the problem statement.
Discussion Status
There is an ongoing exploration of the problem's wording and the assumptions regarding equilibrium. Some participants suggest that the problem may be incorrectly stated, leading to confusion about the forces involved. Guidance has been offered regarding the need for free body diagrams to clarify the forces acting on the block.
Contextual Notes
Participants note potential inconsistencies in the problem statement, particularly regarding the equilibrium of joint B and the implications for the block's motion. The mention of kinetic friction raises questions about the validity of the equilibrium assumption.
- #2
SteamKing
Staff Emeritus
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goldfish9776 said:
Do you feel that the tension has no effect on the block?
The block is in equilibrium as shown. How does the block stay in equilibrium?
- #3
Doc Al
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It's a force acting on the block, so you'd better.goldfish9776 said:
You are told that point B is in equilibrium.goldfish9776 said:
- #4
- 7,201
- 527
tension and frictiongoldfish9776 said:
try a free body diagram of joint B. Please show your work. Is the block in equilibrium?
Homework Equations
The Attempt at a Solution
[/QUOTE]- #5
- #6
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No. But there seems to be a problem in the problem statement, because it states that joint B is in equilibrium which implies that the block must be in equilibrium also. But the solution gives a kinetic (moving) friction value, so something went wrong here, the joint cannot be in equilibrium with the given values.goldfish9776 said:
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- #7
goldfish9776
- 310
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so , the correct one should be 100(9.81)sin20 +150 only ?PhanthomJay said:
- #8
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No, a free body diagram of the block will show that in addition to the two down-the-plane forces you have identified, there is a tension force acting up the plane and possibly a friction force acting parallel to the plane. Now the problem states that the joint B is in equilibrium (not accelerating) and thus the block also must not be accelerating. But when you draw a free body diagram of the joint at B to determine the tension force, and apply that force to the block, there is not enough available static friction force down the plane to keep the block from moving, which is in contradiction with the problem statement. As I see it, the problem is not correctly worded.goldfish9776 said:
- #9
Doc Al
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I agree. A very poorly worded problem. (It reads as if translated from another language.)PhanthomJay said:
You can certainly get the given answer, but you have to disregard (or charitably reinterpret) some of the problem statement. You first assume equilibrium, then ask if that is possible. It's not--as PhanthomJay explains--so the friction must be kinetic.
- #10
goldfish9776
- 310
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isn't it = 100(9.81)sin20 ?PhanthomJay said:
- #11
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No, see Doc Al response above for clarification. If the block is moving, the friction force is just \mu_k(mgcos\theta). You have to charitably reinterpret the problem.goldfish9776 said:
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