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Slope (ramp, inclined plane) problems are usually solved by rotating our frame of reference.
In most problems we are dealing with a horizontal surface. Our "normal" frame of reference specifies vertical and horizontal directions using "up" or "down", "right" or "left" for example. Consider a toboggan pulled along a virtually frictionless icy surface, by a rope at an angle upward. Since the toboggan is moving horizontally, we would analyze the tension on the rope in terms of its horizontal and vertical components.
In problems involving slope the usual approach is to use a different frame of reference. Since motion is parallel to the slope, directions are specified as parallel to the slope or perpendicular to the slope.
Gravity is resolved into two components:
Since normal force in most problems is equal in magnitude to mgcosø, then friction = (µN) = µmgcosø. That is, µmgcosø = friction in most slope problems.
1. A 3.0 m long board has one end raised to a height of 60 cm to form an incline. A 4.0 kg mass is allowed to slide without friction down the entire length of the inclined plane.
a. What is the final speed of the mass when it reaches the bottom ?
b. If the mass is replaced with an 8.0 kg mass, what would be the new speed when it reaches the bottom?
2. A wooden block slides directly down an inclined plane, at a constant velocity of 6 m/s.
a. How large is the coefficient of kinetic friction if the plane makes an angle of 25 degrees with the horizontal?
b. If the angle of incline is changed to 10º, how long far will the block slide before coming to a stop?
3. A 10.0 kg box accelerates at 2.00 m/s2 as it slides down a ramp that makes an angle of 25.0 degrees with the horizontal. Find the coefficient of friction.
4. A 10.0 kg box is pulled up a 45 degree ramp at a constant velocity by a force of 90.0 N acting parallel to the ramp. Find the coefficient of friction.
5. a 10.0 kg box is pulled up a ramp that is at an angle of 20.0 degrees with the horizontal, with the rope that makes and angle of 30.0 degrees with the ramp. if pulling with a force of 70.0 N causes and acceleration of 2.00 m/s2. Find the coefficient of friction.
6. A 10.0 kg box is pulled down a 45 degree ramp by a rope attached to the box that makes a angle of 30.0 degrees with the ramp. If a tension of 10.0 N causes the box to move at a constant velocity, what is the coefficient of friction for the box sliding on the ramp?
7. A car, with a mass of 1100kg, can accelerate on a level road from rest to 21 m/s in 14.0 s. What is the steepest slope this car can climb?
8. 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10 degrees with the horizontal. If there is no friction between the ramp surface and crate, what is the velocity of the crate at the bottom of the ramp?
9. A 10 kg block is placed on top of an inclined plane 10 m long. The coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.2. The inclined plane is 30 degrees from the horizontal. a.) What is the acceleration of the block? b.) When will it hit the ground? c.) What is its impact velocity? d.) What is the angle of repose?
10. An object is being pulled up a 15° incline against a frictional coefficient of 0.15, and requires a force of 835 N parallel to the surface of the ramp to move it at a constant speed. What is the weight of the object?
11. The coefficient of static friction between a box and an inclined plane is 0.35. What is the minimum angle required for the box to begin sliding down the incline?
12. A package slides down a 135 m long ramp with no friction. If the package starts from rest at the top and is to have a speed no faster than 19m/s at the bottom, what should be the maximum angle of inclination?
13. A box of mass (m) is released from rest near the top of a frictionless sphere of radius, R. At what point does the box leave the surface of the sphere?
Selected solutions are printed below.
For solutions to all the problems on this page click here.
Fx and Fy are components of weight, Fg; FN is normal force; Ff is friction
Since Ff = 0, and FN is cancelled by Fy, Fx is the net force
Fnet = Fx
ma = mgsinø
a = gsinø [eqn 1]
a = gsinø
ø = sin -1( .6 / 3.0) = 11.3º
a =(9.8)sin11.3º = 1.92 m/s2
Applying vf2 = vi2+ 2ad gives
vf2 = 0 + 2(1.92)(3.0)
vf = 3.40 m/s
In [eqn 1] we see that the acceleration of the object on the slope (in the absence of friction) does not depend on the object's mass.