A 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 f=6i+8j+20kn is applied on the cube. (use g=10ms2) 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 127 with the positive x-axis in clockwise direction. D : the contact force exerted by the surface on the cube is 1010n.
Answers
Here, normal reaction of the block, R = mg = 50 N.
The friction force on the block is given by,
f = μR = 0.4 × 50 = 20 N
Here the external force is F = 6i + 8j + 20k.
Let F be resolved into two - 6i + 8j and 20k. The block moves over the XY plane due to the component of the F, 6i + 8j.
The magnitude of 6i + 8j force is √(6² + 8²) = 10 N. Since this force is less than the frictional force (10 < 20), the block won't move over the surface.
Hence A is false.
B is false because the frictional force is 20 N.
The component of the external force, 6i + 8j, makes the angle with the positive x axis in anticlockwise direction,
arctan (8 / 6) = 53°
The frictional force is opposite to this force, so it makes an angle 180° - 53° = 127° with the positive x axis in clockwise direction. Hence C is true.
The contact force exerted by the surface on the cube is the normal reaction, i.e., 50 N. Hence D is false.
Explanation:
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