Asked by awoyele tobi
A centreate block mass 25kg is placed on a wooden plant inclined at an angle 32 dergree to the horizontal calculate the force parallel to the inclined plane that would keep the block rest .If the cofficient of fricition between the block and the plane is 0.45
Answers
Answered by
Henry
M*g = 25*9.8 = 245 N. = Wt. of block.
Fp = 245*sin32 = 129.8 N. = Force parallel to the incline.
Fn = 245*Cos32 = 207.8 N. = Normal force.
Fs = u*Fn = 0.45 * 207.8 = 93.5 N. = Force of static friction.
F-Fp-Fs = M*a. a = 0.
F = Ma+Fp+Fs = 0 + 129.8 + 93.5 = 223.3 N.
Fp = 245*sin32 = 129.8 N. = Force parallel to the incline.
Fn = 245*Cos32 = 207.8 N. = Normal force.
Fs = u*Fn = 0.45 * 207.8 = 93.5 N. = Force of static friction.
F-Fp-Fs = M*a. a = 0.
F = Ma+Fp+Fs = 0 + 129.8 + 93.5 = 223.3 N.
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