Total forces acting downward on the plane

[goldfish9776]

Homework Statement

the total force acting downwards the plane = 100(9.81)sin20 + 150= 485.5 N , should I take the tension in B into consideration ? how to get the tension on the rope ?

Homework Equations

The Attempt at a Solution

 

[SteamKing]

Homework Statement

the total force acting downwards the plane = 100(9.81)sin20 + 150= 485.5 N , should I take the tension in B into consideration ?

Do you feel that the tension has no effect on the block?

how to get the tension on the rope ?

The block is in equilibrium as shown. How does the block stay in equilibrium?

 

[Doc Al]

should I take the tension in B into consideration ?

It's a force acting on the block, so you'd better.

how to get the tension on the rope ?

You are told that point B is in equilibrium.

 

[PhanthomJay]

Homework Statement

the total force acting downwards the plane = 100(9.81)sin20 + 150= 485.5 N , should I take the tension in B into consideration ?

tension and friction

how to get the tension on the rope ?.

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]

 

[goldfish9776]

tension and frictiontry a free body diagram of joint B. Please show your work. Is the block in equilibrium?

Homework Equations

The Attempt at a Solution

[/QUOTE]

so , the ttotal forces downwards should be
100(9.81)sin20 +150+196.2 = 681.7N ?

 

[PhanthomJay]

so , the ttotal forces downwards should be
100(9.81)sin20 +150+196.2 = 681.7N ?

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]

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.

so , the correct one should be 100(9.81)sin20 +150 only ?

 

[PhanthomJay]

so , the correct one should be 100(9.81)sin20 +150 only ?

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.

 

[Doc Al]

As I see it, the problem is not correctly worded.

I agree. A very poorly worded problem. (It reads as if translated from another language.)

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.

 

[goldfish9776]

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.

isn't it = 100(9.81)sin20 ?

 

[PhanthomJay]

isn't it = 100(9.81)sin20 ?

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.