Solving the Acceleration of a Bob Attached to a Train

AI Thread Summary
The discussion focuses on solving a physics problem involving a bob attached to a train that is accelerating. Participants suggest using Newton's second law and analyzing the forces acting on the bob, including weight and tension. A free body diagram is recommended to visualize the forces, and there is emphasis on separating horizontal and vertical components for analysis. The conversation also touches on equilibrium conditions and moments, with participants confirming the validity of equations derived from these principles. The overall goal is to demonstrate that the train's acceleration is 2.63 m/s².
jinhuit95
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Homework Statement


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.


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! :)
 
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Use Newton's 2nd law. What forces act on the bob? Analyze horizontal and vertical components.
 
How do you know what to use?? I was thinking of using mg sin θ in the first place.
 
Start by drawing a free body diagram of the bob showing all forces acting on it. What forces act on the bob?
 
Weight and tension of the string!
 
jinhuit95 said:
Weight and tension of the string!
Right!

Now apply Newton's 2nd law. I suggest analyzing horizontal and vertical components separately.
 
Alright I'll try! Can I ask you another question over here??
 
jinhuit95 said:
Can I ask you another question over here??
Ask away.
 
http://via.me/-2ostlh4
http://via.me/-2ostlh4
Here's the question and I did part of it and I don't know how to continue!
 
Last edited by a moderator:
  • #10
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
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
jinhuit95 said:
Other condition will be the upward forces = downward forces?? Am I correct??
Yes.
My working was W*0.5L cos θ = Y*L sin θ + f* L cos θ and that frictional force cause a moment too??
Looks good.
 
  • #13
Doc Al said:
Yes.

Looks good.

so that means W= X+f?? And I just subt w into the above equation I did??
 
  • #14
jinhuit95 said:
so that means W= X+f?? And I just subt w into the above equation I did??
Yep.
 
  • #15
X=W
X=W/2 + W/2
Find value of W/2 in term of Y.
 
  • #16
azizlwl said:
X=W
Don't neglect the friction force from the wall.
 
  • #17
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
azizlwl said:
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.
But that's not the problem being discussed here. The wall is rough.
 
  • #19
Yes i make a mistake there.
With or without friction the equation still valid.
 
  • #20
azizlwl said:
With or without friction the equation still valid.
How can you say that?
 

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