Asked by Adarsh
what force parallel to an inclined plane of height 3 meter and base 4 meter can just support a block mass 500 gram on the plane ? the coefficient of friction 0.25
Answers
Answered by
MathMate
Given:
θ=arctan(3/4)=36.87° approx.
m=0.5 kg
μ=0.25
weight of block can be resolved into
force down plane, F = mg sin(θ)
and normal force, N = mg cos(θ)
Let P be applied force parallel to plane
(upwards) to maintain equilibrium, then
Consider equilibrium along plane:
F=P+μN
mg sin(θ)=μmg cos(θ)+P
solve for P.
θ=arctan(3/4)=36.87° approx.
m=0.5 kg
μ=0.25
weight of block can be resolved into
force down plane, F = mg sin(θ)
and normal force, N = mg cos(θ)
Let P be applied force parallel to plane
(upwards) to maintain equilibrium, then
Consider equilibrium along plane:
F=P+μN
mg sin(θ)=μmg cos(θ)+P
solve for P.
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