Frictionless Force question

  • Thread starter aliciaw0
  • Start date
  • #1
aliciaw0
17
0
Find an expression for the magnitude of the horizontal force F in the figure for which m1 does not slip either up or down along the wedge. All surfaces are frictionless.

since it has nothing to do with the friction i am thinking that m1 and the force have to have the same acceleration in the horizontal direction for it not to slide up or down, but im not sure what to do from there?

thanks
 

Attachments

  • phy.jpg
    phy.jpg
    8.3 KB · Views: 1,603

Answers and Replies

  • #2
abercrombiems02
114
0
Ok this is what i recommend you do. Setup and FBD for the smaller block alone. You should get your sum of forces vectically to equal zero. The horizontal sum of forces on the smaller block must equal its mass multiplied by its acceleration (F=ma). You know the blocks must accelerate together so acc_block = acc_system. Now apply sum(Fext) = m_sys*a_sys.
 
  • #3
aliciaw0
17
0
thanks i tried what you told me too and i think i still might have done it wrong?

f(block1)y: Fn-Fgcos(theta)=0

F(block1)x: Fgsin(theta)=m1a Fgsin(theta)/m1= a

system x: F= (m1+m2) ((m1g*sin(theta)/m1) and they said my answer was wrong because it doesnt depend on theta. so if you know where to go from here it would be helpful! thanksss
 
  • #4
Doc Al
Mentor
45,410
1,845
aliciaw0 said:
f(block1)y: Fn-Fgcos(theta)=0

F(block1)x: Fgsin(theta)=m1a
The acceleration of the smaller block is purely horizontal, so define your components accordingly. Let the x-axis be horizontal, not down the plane; let the y-axis be vertical, not perpendicular to the plane.

Redo these two equations. The normal force will have both vertical and horizontal components. The weight should be set equal to [itex]m_1 g[/itex].
 
  • #5
aliciaw0
17
0
yay i got it right thankkk you

Y: Fn= (m1*g)/cos(theta)

X: (m1*g*tan(theta))/m1 = a

system: F= (m1+m2) ((m1*g*tan(theta))/m1 ) =]
 

Suggested for: Frictionless Force question

  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
2
Views
9K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
28
Views
4K
Replies
4
Views
1K
Replies
4
Views
2K
Replies
6
Views
4K
  • Last Post
Replies
1
Views
2K
Replies
2
Views
2K
Replies
2
Views
2K
Top