# Homework Help: Bob of mass

1. Jul 3, 2012

### jinhuit95

1. The problem statement, all variables and given/known data
A bob of mass 0.35 kg is attached by a string to the roof of a train that is accelerating rightwards and the mass is inclined at an angle of 15° to the left. Show that the acceleration of the train is 2.63 ms^-2.

3. The attempt at a solution
I don't know how to exactly start doing this because I don't know what concepts I should use here. Please help! :)

2. Jul 3, 2012

### Staff: Mentor

Use Newton's 2nd law. What forces act on the bob? Analyze horizontal and vertical components.

3. Jul 3, 2012

### jinhuit95

How do you know what to use?? I was thinking of using mg sin θ in the first place.

4. Jul 3, 2012

### Staff: Mentor

Start by drawing a free body diagram of the bob showing all forces acting on it. What forces act on the bob?

5. Jul 3, 2012

### jinhuit95

Weight and tension of the string!

6. Jul 3, 2012

### Staff: Mentor

Right!

Now apply Newton's 2nd law. I suggest analyzing horizontal and vertical components separately.

7. Jul 3, 2012

### jinhuit95

Alright I'll try! Can I ask you another question over here??

8. Jul 3, 2012

9. Jul 3, 2012

### jinhuit95

http://via.me/-2ostlh4
http://via.me/-2ostlh4 [Broken]
Here's the question and I did part of it and I don't know how to continue!

Last edited by a moderator: May 6, 2017
10. Jul 3, 2012

### Staff: Mentor

You took moments about the bottom end of the ladder, which is good. What about the other conditions for equilibrium?

(It's a bit difficult to read your work from the diagram, so write it out here if you want further critique.)

11. Jul 3, 2012

### jinhuit95

Other condition will be the upward forces = downward forces?? Am I correct??
My working was W*0.5L cos θ = Y*L sin θ + f* L cos θ and that frictional force cause a moment too??

12. Jul 3, 2012

### Staff: Mentor

Yes.
Looks good.

13. Jul 3, 2012

### jinhuit95

so that means W= X+f?? And I just subt w into the above equation I did??

14. Jul 3, 2012

### Staff: Mentor

Yep.

15. Jul 3, 2012

### azizlwl

X=W
X=W/2 + W/2
Find value of W/2 in term of Y.

16. Jul 3, 2012

### Staff: Mentor

Don't neglect the friction force from the wall.

17. Jul 3, 2012

### azizlwl

Taking as No friction on the wall,
If we take moment at the base,
(L/2)WCosθ=LYSinθ
W/2=YTanθ.

Then we can proof the equation.

18. Jul 3, 2012

### Staff: Mentor

But that's not the problem being discussed here. The wall is rough.

19. Jul 3, 2012

### azizlwl

Yes i make a mistake there.
With or without friction the equation still valid.

20. Jul 3, 2012

### Staff: Mentor

How can you say that?