xMerlina
Dec 9, 2010, 08:19 PM
5. A box weighing 450 N is pulled along a level floor at constant speed by a rope that makes an angle of 30ΒΊ with the floor. If the force on the rope is 260 N, what is the coefficient of friction?
Answer: 0.703
I keep getting .577
Unknown008
Dec 10, 2010, 02:17 AM
Resolve all the forces acting on the box.
Along the vertical, you have the normal force and the component of the rope which equals the weight of the box.
Hence, you get:
mg = R + F\sin(\theta)
450 = R + 260\sin(30)
Along the horizontal, since the box is moving at a constant speed, the force of friction equals the component of the pulling force.
F_f = \mu R
\mu R = F\cos(\theta)
\mu R = 260\cos(30)
You have to find mu, can you do it? :)
xMerlina
Dec 10, 2010, 03:38 PM
Ohh, that makes more sense. Thanks! ;)