Asked by Jat
A particle of mass m can just rest on a rough plane inclined at 30 degree to the horizontal without slipping down. show that the least horizontal force needed to maintain its position if the inclination is increased to 45 degree is 0.268mg.
Answers
Answered by
drwls
From the first bit of information, the static friction coefficient (Us) is given by
M*g*cos30*Us = M*g*sin30
Us = tan30 = 0.5773
At a 45 degree tilt with that static friction coefficient, the required horizontal force to prevent slipping is given by:
F cos45+ (M*g cos45+Fsin45)*(0.5773) =
= M*g*sin45
F(0.7071 +0.4082) = M*g(.7071 -.4082)
F = .2989/1.115 Mg = 0.268 M g
I have assumed that the applied frorce in horizontal, not up and along the incline. That would lead to a different answer.
M*g*cos30*Us = M*g*sin30
Us = tan30 = 0.5773
At a 45 degree tilt with that static friction coefficient, the required horizontal force to prevent slipping is given by:
F cos45+ (M*g cos45+Fsin45)*(0.5773) =
= M*g*sin45
F(0.7071 +0.4082) = M*g(.7071 -.4082)
F = .2989/1.115 Mg = 0.268 M g
I have assumed that the applied frorce in horizontal, not up and along the incline. That would lead to a different answer.
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