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Find the intial velocity given the coefficient of kinetic friction

  • Thread starter ScullyX51
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  • #1
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Homework Statement



You and your friend Peter are putting new shingles on a roof pitched at 21^\circ . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.4 m away, asks you for the box of nails. Rather than carry the 2.4 kg box of nails down to Peter, you decide to give the box a push and have it slide down to him.
If the coefficient of kinetic friction between the box and the roof is 0.52, with what speed should you push the box to have it gently come to rest right at the edge of the roof?

Homework Equations


f=ma
f(kinetic friction)= NMk


The Attempt at a Solution


Ok..here is what I did:
I drew a FBD of the mass, and got the following:
mg pointing straight down in negative g direction, N diagnaly up to the left, and Mk diagnaly up to the right
Then I got there components as follows ( I think I may have used the wrong angles):
for the normal force: N(cos45i-sin45j)
Fgrav: -mgj
F(friction): MkN(-cos 69+sin69)
( I used 69 for the angle because I made a right triangle from the roof at angle 21, and then subtracted 21 from 90 to get the angle to get the components of friction.)

then for forces in the i direction I have: Ncos45-MkN69=ma
a= Ncos45-Mkn69/m

for forces in the j direction I got:
MkNsin69-Nsin45-mg=0
MkN-Nsin45=mg
N(Mksin69-sin45)=mg
N=mg/Mksin69-sin45
N=(2.4)(9.8)/.52sin69-sin45
N= -106

I plugged N into the equation for a and got:
-106(cos45)-.52(-106)cos69/2.4
a=-14.73

Then to find the initial velocity I used the kinematics equation: V^2= V(i)^2+2a(delta)x
0=V(i)^2+2(-14.73)(5.4)
V(i)+ (159.084)^1/2
V(i)= 12.61 m/s

Hopefully this isn't to hard to follow...and someone can tell me where I went wrong!
 

Answers and Replies

  • #2
108
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Then I got there components as follows ( I think I may have used the wrong angles):
for the normal force: N(cos45i-sin45j)
This 45 comes up a fair bit throughout the solution. Was there a reason you decided to use that angle - was it one of the angles on your free body diagram / sketch?
 
  • #3
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It doesn't really make sense where I got it from. From my FBD the way I drew N diagonally up to the left, it looked like it was cutting through a 90 angle. I am just really confused on what to do with the given angle (21 degrees), since I'm picturing that as being the "point" of the roof I can see how to incorporate that into my FBD
 
  • #4
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On re-reading, I think it's a bit unfair of the question to expect you to know what the pitch of a roof is defined to be (unless there's a picture to go with it) - though a 21 degree angle with the horizontal feels more practical than a 21 degree angle with the vertical, or a 21 degree angle between the two halves of the roof!

I believe that in this case, it means that the roof makes an angle of 21 degrees with the horizontal (though this is not how the pitch of a roof is normally quoted from what I can tell from a quick Google - see http://books.google.co.uk/books?id=...X&oi=book_result&resnum=8&ct=result#PPA209,M1 for example).

Hopefully, that will help with your free body diagrams, angles of forces, components, etc. Though bear in mind I only figured out what I think "pitch" means from a quick web search, and it may be as well for you to try to find a decent definition of it in case I'm wrong!
 

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