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?
Answers
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:
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: