Laws of MotionEasy
Question
solid cube of mass 5 kg is placed on a rough horizontal surface, in xy-plane as shown. The friction coefficient between the surface and the cube is 0.4. An external force 
N is applied on the cube.
N is applied on the cube.Options
A.The block starts slipping over the surface
B.The friction force on the cube by the surface is 10 N.
C.The friction force acts in xy-plane at angle 127o with the positive x-axis in clockwise direction.
D.The contact force exerted by the surface on the cube is 10
N.
N.Solution
Normal force = 5g - 20 = 30 N
fmax = μN = 0.4 × 30 = 12N
∵ |6î + 8ĵ| = 10N
∵ fmax > 10N ⇒ f = 10N ⇒ static friction
fmax = μN = 0.4 × 30 = 12N
∵ |6î + 8ĵ| = 10N
∵ fmax > 10N ⇒ f = 10N ⇒ static friction
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