Please help to find mu, frictional force and normal force

AI Thread Summary
To find the coefficient of friction (μ) for a 5 kg mass on a board tilted at 30 degrees, the normal force (Fn) is calculated as 42.47 N. The weight components are determined as (Fw)x = 24.52 N and (Fw)y = 42.47 N. The frictional force (Ff) equals the force down the ramp, which is (Fw)x at the angle of 30 degrees. By using the relationship μ = Ff/Fn, the coefficient of friction can be derived. The discussion emphasizes the importance of breaking forces into components to solve for μ effectively.
jaybabbar
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a body of 5 kg mass is resting on a board which is gradually tilted until, at an angle of 30 degrees to the horizontal, the body begins to move down the plane. find coefficient of friction and magnitude of the normal and frictional forces when the body begins to slip.

i managed to workout Force due to weight i.e. (Fw)x = 24.52N and (Fw)y = 42.47N also the Fn(Normal force) = 42.47 but i can't workout mu(coefficient of friction) as i don't have Ff(friction force) from the formula mu = Ff/Fn
Thanks
 
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welcome to pf!

hi jaybabbar! welcome to pf! :smile:

(have a mu: µ and try using the X2 icon just above the Reply box :wink:)
jaybabbar said:
… (Fw)x = 24.52N and (Fw)y = 42.47N also the Fn(Normal force) = 42.47 but i can't workout mu(coefficient of friction) as i don't have Ff(friction force) from the formula mu = Ff/Fn

ah, but you know that the components of force in the x direction must add to zero …

so you know that Ff = (Fw)x :wink:
 
The body moves at 30 degrees so you need to use trig to break it into components at that angle. The force going down the ramp must equal the frictional force at the point friction breaks. The force going into the ramp is equal to normal and using both you can find coefficient of friction.
 
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