1. Break the pulling force in to a vertical and horizontal component. The vertical component subtracts from the normal weight. So figure friction based on that.
Net horizontal force=m*a
horizonalPulling-friction= ma
sove force.
2. Again, break the pulling into a normal component and a parallel(plane) component.
the normal force will be mg*cosTeta-normalofPullingcomponent.
Friction= mu*normal force above.
Component of gravity down the plane, mgSinTheta..
Net up force=ma
ParallelPulling-friction-gravitydownplane=ma
solve for a
1.A student moves a box of books down the hall
by pulling on a rope attached to the box. The
student pulls with a force of 181 N at an angle
of 20.8◦ above the horizontal. The box has a
mass of 40.3 kg, and μk between the box and
the floor is 0.29.
The acceleration of gravity is 9.81 m/s2 .
Find the acceleration of the box.
2.Now the student moves the box up a ramp
(with the same coefficient of friction) inclined
at 10.3◦ with the horizontal.
b) If the box starts from rest at the bottom
of the ramp and is pulled at an angle of 20.8◦
with respect to the incline and with the same
181 N force, what is the acceleration up the
ramp?
Answer in units of m/s2.
2 answers
Fx = 181 cos 20.8
normal force = m g - 181 sin 20.8
friction force =.29(mg-181 sin 20.8)
ma= 181cos 20.8-.29(mg-181 sin 20.8)
divide by m to get a
now you do the second one the same way
normal force = m g - 181 sin 20.8
friction force =.29(mg-181 sin 20.8)
ma= 181cos 20.8-.29(mg-181 sin 20.8)
divide by m to get a
now you do the second one the same way
2 answers