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This is pissing me off! I cant solve this simple Physics question
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| stk |
A loaded penguin sled weighing 80 N rests on a plane inclined at 20° to the horizontal (Fig. 6-28). Between the sled and the plane the coefficient of static friction is 0.28, and the coefficient of kinetic friction is 0.14.
(a) What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane?
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this is how i tried to solve it
note *@=theta, F=Force of magnitude, f= Friction, N = Normal Force, mg= Force of Gravity, u = mew (friction coefficient)
What i did was i drew a free body diagram and used sum of forces
On the X coordinate i got F-mgsin@-f=0
On the Y N=mgcos@
After summing the two equations, i derived this:
F=mgsin@+umgcos@
Then i plugged in all the variables (i used the static coefficient friction) and got 48.1444, which is not the answer...Can anyone please tell me what im doing wrong? Thanks |
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| DjJade |
On the X coordinate you need F-mgsin@+f=0
the question wants to know the force that you need to keep the sled from slipping. f always opposes mgsin@ so you want F to be in the same direction as f to help f...to keep the sled from sliding down.
make sure you use the coefficient for static friction since it will be at rest. |
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| whiskers |
hope these help
/bored |
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| stk |
ok so i got F+f-fgsin@=0
so F=Fgsin@-f
and f=uN
and N = mgcos@
so F=Fgsin@-umgcos@
so F=27.36- (.28)(80)cos(20) = 6.31 N
Can anyone verify that this is right? Cause i only have 1 more submission left before this gets graded
thanks |
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| whiskers |
i get
80sin(20)-0.28*80cos(20) = 6.31249676 N ;) |
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| {b.s.e.} |
| Koolaide man sez, I don't have physics, but I do have vodka. |
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| jinxed84 |
| the sad thing is i learned how to do that in HS and i forgot how to do it now. shows how much that was worth. |
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