Horizontal force kinematics help

  • Thread starter frozen7
  • Start date
  • #1
frozen7
163
0
The figure shows a small object of mass m = 1kg can slide without friction on a wedge of mass M = 3kg inclined at angle [tex]\delta[/tex]=30 degree . What horizontal force F must be applied to the wedge if the small block is not to move with respect to the wedge?

I do it in this way:

By Newton`s 3rd law: the small object will exert a same force F to the wedge, so

mgsin30 = Fcos30
F = 5.66N

But my answer is wrong. Can anyone teach me how to solve it?
Thanks.
 

Attachments

  • untitled.JPG
    untitled.JPG
    3.7 KB · Views: 354

Answers and Replies

  • #2
stunner5000pt
1,455
2
mg sin 30 = F
no Fcos30 coz you're applying this force in the same direction as the slant.
 
  • #3
Fermat
Homework Helper
872
1
I've got an idea for the solution of this problem, but I'm not sure how feasible it is :frown:

The small mass is moving on a smooth surface, so there is no friction.
The accelerating force down the slope is mg.sin@ (where @ is the slope of the wedge). So the acceln of the mass down the slope is a=g.sin@.
The mass has a horizontal acceleration of a_h = a.cos@ = g.sin@.cos@ = ½g.sin(2@).

My idea: If the wedge is now moved forward with a horizontal acceleration of a_h, equal to that of the small mass, then this will provide vertical support for the small mass such that no vertical movement of it will happen and it will be motionless wrt the wedge.

Any comments on my idea ??

Assuming my idea is correct,

F = (m+M)a_h
F = (m+M).½g.sin(2@).
F = ½g.(m+M).sin(2@).
=================

F = 4.9*(1 + 3)*sin(60)
F = 16.974 N
==========
 
Last edited:
  • #4
frozen7
163
0
There are 4 choices answers for this question:
a)22.6
b)28.4
c)32.5
d)36.2
 
  • #5
mukundpa
Homework Helper
524
3
If it is not sliding m is also accelerating with the same acceleration.

Draw a free body diagram of object m resolve the forces in horizontal and vertical direction and you will get acceleration probably g tan(theta).
 
  • #6
frozen7
163
0
But, is this relation correct?

mgsin30 = Fcos30

Furthermore, can you explain to me about the free body diagram? I can't see why a = g tan(theta)

Thanks.
 
  • #7
mukundpa
Homework Helper
524
3
Free body diagram mens the diagram showing all forces acting on a single body. For m it is in the attaqchment.
 

Attachments

  • incline.doc
    19 KB · Views: 142
  • #8
mukundpa
Homework Helper
524
3
As m is not in equlibrium N is not equal to mg cos@.
 
  • #9
frozen7
163
0
Mukundpa...Thanks a LOT...
 
  • #10
mukundpa
Homework Helper
524
3
You are always Welcome :smile: :smile:
 

Suggested for: Horizontal force kinematics help

Replies
5
Views
324
Replies
1
Views
625
Replies
3
Views
717
Replies
15
Views
633
Replies
8
Views
388
  • Last Post
Replies
8
Views
590
Replies
13
Views
380
Replies
13
Views
338
Replies
3
Views
379
Top