Kinetic friction

  • Thread starter Pencil
  • Start date
  • #1
17
0
To move a large crate across a rough floor, you push down on it at an angle of 21°. Find the force necessary to start the crate moving, given that the mass of the crate is m = 32 kg and the coefficient of static friction between the crate and the floor is 0.57.

N=mg=(32)(9.81)= 313.92 * coefficaint of mew sub s=.57=178.9
Ok, I think I did that right....only I'm not sure what to do with the 21° angle. I did 178.9cos(21)=167.0 but that's not right.
 

Answers and Replies

  • #2
486
1
Did you draw a diagram with the applied force resolved into x and y components? (BTW, that's where the angle comes in)
Because in this case, N does not equal mg.
 
  • #3
17
0
Like this?
Ny=N
Fky=0
Wy=-mgsin(21
Nx=0
Fkx=-mewk
Wx=0
I did Fk=mewkN=.57*(mgsin(21)=.57*112.5=64.12 I did somthing wrong...again...but I don't know what...
 
  • #4
486
1
where's the force of the push? It has both an x and a y component.
But yeah, that's the sort of chart I had in mind. :grin:
 

Related Threads on Kinetic friction

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
2
Replies
26
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
17K
Top