Draw the free body diagram which reveals the net forces on the body.
The force F at 20 degrees with horizontal has an upward component of Fsin(20) which has the effect of reducing the normal reaction N from mg to mg-Fsin(20).
The horizontal component of Fcos(20) is the only force that resists friction, so may be equated with the frictional force of μN.
So equating horizontal forces,
μN=Fcos(20), or
μ(mg-Fsin(20))=Fcos(20).
Solve for μ.
A body with a mass of 490kg is pulled along a horizontal plane at a constant velocity.the pulling force is 1.2 kN is making an angle of 20degrees with the plane.
A.determine the coefficient of friction
3 answers
Coefficient of friction is 13.0? Am I correct?
I do not see how you can get 13.0.
Please show your work (must use consistent units, kg, N, s, m)
F=1.2kN=1200N
m=490 kg
g=9.8 or 9.81 m/s^2
Perhaps it was a mistake on the calcultor, so please try again.
μ should be between 0 and 1.
Please show your work (must use consistent units, kg, N, s, m)
F=1.2kN=1200N
m=490 kg
g=9.8 or 9.81 m/s^2
Perhaps it was a mistake on the calcultor, so please try again.
μ should be between 0 and 1.
1 answer