Question

A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3N. then the magnitude of contact force is?​

Answer 1

Answer:

The Normal Reaction would be given by:\brSpace N=mg=0.400\times 10=4 N\brSpace The frictional force is given by:\brSpace f=3 N\brSpace Hence the angle made by the contact force would be given by:\brSpace \theta =tan^{-1}\left (\frac{3}{4} \right )=37^{0}\brSpace and the magnitude of the contact force is given by:\brSpace F=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=5 N

Explanation:

Answer 2

Answer:

The Normal Reaction would be given by:\brSpace N=mg=0.400\times 10=4 N\brSpace The frictional force is given by:\brSpace f=3 N\brSpace Hence the angle made by the contact force would be given by:\brSpace \theta =tan^{-1}\left (\frac{3}{4} \right )=37^{0}\brSpace and the magnitude of the contact force is given by:\brSpace F=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=5 N

Explanation: