Object on an incline with friction

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An object of mass 3 kg is on a 45-degree inclined plane with a coefficient of friction of 0.2 and a 10 N force acting at a 30-degree angle. The calculated normal reaction force is 26.2 N, leading to a friction force of 5.24 N, but confusion arises regarding the angle's reference point. The angle of 30 degrees is measured relative to the object itself, not the horizontal or the plane. Corrections indicate a sign error in the normal reaction calculation and suggest using sine instead of cosine for the angle. Understanding the normal force conceptually is crucial for accurate calculations.
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An object of mass 3 kg lies on a rough plane inclined at 45 degree to the horizontal. The coefficient of friction is 0.2 and a force of 10 N acts at 30 degree to the object. Calculate the friction. According to the booklet the answer is 3.24 N

My work

F = u R


Normal reaction = 10 sin 30 + 30 cos 45 = 26.2

U = 0.2 x 26.2 = 5.24 N

I am confused HELPPPPPP.. where is the fault, is there something that i have not considered
 
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What is the 30 degrees measured relative to? Is it relative to the plane the block is sitting on or relative to the horizontal?
 


w3390 said:
What is the 30 degrees measured relative to? Is it relative to the plane the block is sitting on or relative to the horizontal?


It is relative to the block(object)
 


So you have a block sitting on a slope. Is the block moving?
 


Your normal reaction has a sign error. Also that cos 45 should be sin 45. I know numerically they are the same thing, but for what it actually represents, it should be sin.

I'll give you a hint, think about what the normal force is conceptually, not mathematically
 
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