A particle of mass m rests on a rough plane which is inclined at an angle θ to the horizontal. The coefficient of friction between the plane and the particle is µ.
A horizontal force of magnitude P is applied to the particle and is just sufficient to prevent the particle from sliding down the slope.
If however a force of magnitude 2P is applied parallel to the slope and acting up the slope, then the particle is just on the point of slipping up the slope.
If instead a particle of mass 2m sits on the slope then a force 2P up the slope is just sufficient to prevent the particle from sliding down.
Find the values of θ, µ and P in terms of m and g where g is the acceleration due to gravity.
I got these 3 equations
Pcosθ+µmgcosθ+µPsinθ=mgsinθ
2P=µmgcosθ+mgsinθ
2P+2µmgcosθ=2mgsinθ
solving them I get
P= (root 3 over 3)mg
θ=60 degrees
µ= root 3 over 3
Have I gone wrong somewhere? How do I get µ and θ in terms of mg if mg always gets cancelled?
A horizontal force of magnitude P is applied to the particle and is just sufficient to prevent the particle from sliding down the slope.
If however a force of magnitude 2P is applied parallel to the slope and acting up the slope, then the particle is just on the point of slipping up the slope.
If instead a particle of mass 2m sits on the slope then a force 2P up the slope is just sufficient to prevent the particle from sliding down.
Find the values of θ, µ and P in terms of m and g where g is the acceleration due to gravity.
I got these 3 equations
Pcosθ+µmgcosθ+µPsinθ=mgsinθ
2P=µmgcosθ+mgsinθ
2P+2µmgcosθ=2mgsinθ
solving them I get
P= (root 3 over 3)mg
θ=60 degrees
µ= root 3 over 3
Have I gone wrong somewhere? How do I get µ and θ in terms of mg if mg always gets cancelled?
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