Resolving alpha with friction and angles

  • Thread starter bob4000
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  • #1
bob4000
40
0
a straight footpath makes an angle of aplha with the horizontal. an object P of weight 1250N rests on footpath. the coefficient of friction is 0.1. the least magnitude of a force, acting up the footpath, which will hold the object at rest on the footpath is 50N. by treating the object as a particle show that the value of alpha satisfies 10sinaplha-cosalpha=0.4

i have drawn FBD and labbelled all of the forces and tried to solve for alpha using simultaneous equations and other methods (the names of which I am not too sure). could anyone please tell me what to do next?

much obliged
 

Answers and Replies

  • #2
Nylex
552
2
Post your work, so we can see exactly where you are.
 
  • #3
bob4000
40
0
f=ma parallel to slope

50-F-1250sinaplha=0

f=ma perpendicular to slope

N-1250cosalpha=0

F=mu*N
mu=0.1
N=F/mu;N=10F
; 10F-1250cosalpha=0
F+1250sinalpha=50

ps, I am half asleep so excuse any blatant stupidity
 

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