# Forces in equilibrium

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1. Feb 12, 2017

### Faiq

1. The problem statement, all variables and given/known data

[moderator note: Post edited to make problem statement images visible]

https://www.physicsforums.com/attachments/question-jpg.113056/

My question is when forces are resolved horizontally, why is the friction force at A not taken into account?

3. The attempt at a solution
Find the value of u (in terms of W)

Note:- The W next to u in the picture is actually Reaction at D, R(d)

Given solution:

In the given solution, why is the friction force at A not taken into account?

Last edited by a moderator: Feb 12, 2017
2. Feb 12, 2017

### Staff: Mentor

Moderator note: Thread cleaned up after amending the original post to clarify the problem.

3. Feb 12, 2017

### Faiq

I just want to know when they resolved the forces horizontally, why does their answer makes sense even though they didnt include the frictional force at point A

4. Feb 12, 2017

### Staff: Mentor

If you read the problem statement (which is now visible ) , how is the rod joined to the plane at point A?

5. Feb 12, 2017

### Faiq

By a hinge

6. Feb 12, 2017

### Staff: Mentor

Right. The hinge will be attached (probably bolted) to the floor. No sliding friction involved, the rod is firmly attached.

7. Feb 12, 2017

### Faiq

Oh but there will still be a reaction at the hinge and we have to take account the horizontal component of that

8. Feb 12, 2017

### Staff: Mentor

Since it is fixed its horizontal reaction will be equal to and opposite of any horizontal force applied to it.

My interpretation of the shown solution is that they were concentrating on the block's FBD, where the local friction has to counter the other external forces applied to the block.

9. Feb 12, 2017

### Faiq

So even if there was a resultant horizontal force at A, it wont effect the block?

10. Feb 12, 2017

### Staff: Mentor

Not if the system is in equilibrium. Only forces applied directly to the block affect the block.

11. Feb 12, 2017

### Faiq

But the force at A will effect the rod which will effect the block

12. Feb 12, 2017

### Staff: Mentor

The rod does not move. Nothing moves. How can the force at A affect how the rod presses on the block? That was determined by taking moments about A.

It is enough to know that the rod cannot slip at A.

13. Feb 12, 2017

### Faiq

oh Okay thank you. Can you also help me in another question

14. Feb 12, 2017

### Staff: Mentor

Start a new thread for it and I or someone else will be happy to help.