A 12 Kg box is put on the surface of an inclined plane at 38 ° with the horizontal. The coefficient of kinetic friction is 0.35. Determine the magnitude of the force exerted by the inclined plane on the box.
Given:
"m=12\\:\\rm kg"
"\\theta=38^{\\circ}"
"\\mu=0.35"
The friction force
"F=\\mu N=\\mu mg\\cos\\theta""F=0.35*12*9.8*\\cos38^{\\circ}=32.4\\:\\rm N"
Comments
Leave a comment