A minimum force F is applied to a block of mass 102 kg toprevent it from sliding on a plane with an inclination angle30° with the horizontal. If the coefficients of static andkinetic friction between the block and the plane are 0.4and 0.3 respectively, then the force F is:
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Given A minimum force F is applied to a block of mass 102 kg to prevent it from sliding on a plane with an inclination angle 30° with the horizontal. If the coefficients of static and kinetic friction between the block and the plane are 0.4 and 0.3 respectively, then the force F is:
We know that
F = f (applied) – f(gravity) – f(friction)
= fa – fg + ff
So for equilibrium Fn = 0 and along the inclined plane we have mg sin θ
Fa = Fg – Ff
Fa = mg sin θ – μ s x N
= mg sin θ – μs x mg cos θ
= mg (sin θ – μs cos θ)
= 102 x 10 (sin 30 – 0.4 cos 30)
= 102 x 10 x 0.154
= 157 N
I really hope this helps!
Explanation:
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