A body of weight 300 N is lying on a rough horizontal plane having a coefficient of frict ion as 0.3. What would be the magnitude of the force, which can move the body, while acting at an angle of 25 with the horizontal? a)67 N b)77 N c)87 N d)97 N
Answer (1)
30th Jan, 2022
Hi there,
Answer is C -87 N
as it is given weight=300 N
u=0.3
Let p is force and force of friction is f
resolving forces horizontally
F=p cos 25
=p×0.9063
Vertically it is R= W-p sin 25 degree
R= 300-P × 0.4226
Force of friction
0.0963×p=uR= 0.3×(300-0.4226P)
=90-0.1268p
=0.9063p+0.1268p= 90
(1.0331)p=90
P=87.1 N
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