Resolving alpha with friction and angles

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SUMMARY

The discussion focuses on resolving the angle alpha of a footpath inclined to the horizontal, where an object weighing 1250N is at rest. The coefficient of friction is given as 0.1, and it is established that the least force required to keep the object at rest is 50N. By applying the equations of motion and friction, the relationship 10sin(alpha) - cos(alpha) = 0.4 is derived, indicating the necessary conditions for equilibrium.

PREREQUISITES
  • Understanding of free body diagrams (FBD)
  • Basic knowledge of Newton's laws of motion
  • Friction concepts, specifically static friction and coefficients
  • Trigonometric functions related to angles and forces
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  • Study the derivation of forces in inclined planes using Newton's second law
  • Learn about static friction and its role in equilibrium problems
  • Explore trigonometric identities and their applications in physics
  • Practice solving problems involving forces on inclined surfaces
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bob4000
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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 I am not too sure). could anyone please tell me what to do next?

much obliged
 
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Post your work, so we can see exactly where you are.
 
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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