Resolving alpha with friction and angles

  • Thread starter bob4000
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  • #1
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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 im not too sure). could anyone please tell me what to do next?

much obliged
 

Answers and Replies

  • #2
551
1
Post your work, so we can see exactly where you are.
 
  • #3
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, im half asleep so excuse any blatant stupidity
 

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