Practice 2.2B
Inclined Plane Problems
A small box is on a ramp tilted at an angle θ above the horizontal. The box may be subject to the
following forces: frictional (f ) ,gravitational (mg), pulling or pushing (FP) and normal (I). In the following
free-body diagrams for the box, the lengths of the vectors are proportional to the magnitudes of the
forces.
1. Which figure best represents the free-body diagram for the box if it is accelerating up the ramp?
(A) Figure A (B) Figure B (C) Figure C (D) Figure D (E) Figure E
The normal force must point perpendicular to the surface and the weight must point down.
In order to accelerate up the ramp, there must be an applied force up the ramp. If the box is
accelerating up the ramp, friction acts down the ramp, opposite the motion.
2. Which figure best represents the free-body diagram for the box if it is at rest on the ramp?
(A) Figure A (B) Figure B (C) Figure C (D) Figure D (E) Figure E
The normal force must point perpendicular to the surface and the weight must point down. If
the box is at rest on the ramp, friction acts up the ramp, opposing the tendency to slide down
3. Which figure best represents the free-body diagram for the box if it is sliding down the ramp at
constant speed?
(A) Figure A (B) Figure B (C) Figure C (D) Figure D (E) Figure E
The normal force must point perpendicular to the surface and the weight must point down. If
the box is sliding down at constant speed, friction acts up the ramp, opposing the motion
4. A crate of toys remains at rest on a sleigh as the sleigh is pulled up a hill with an increasing
speed. The crate is not fastened down to the sleigh. What force is responsible for the crate’s
increase in speed up the hill?
(A) the contact force (normal force) of the ground on the sleigh
(B) the force of static friction of the sleigh on the crate
(C) the contact force (normal force) of the sleigh on the crate
(D) the gravitational force acting on the sleigh
(E) no force is needed
The only force in the direction of the crate’s acceleration is the force of friction from the sleigh
5.
A force F is used to hold a block of mass m on an incline as shown in the diagram (see above). The plane
makes an angle of θ with the horizontal and F is perpendicular to the plane. The coefficient of friction
between the plane and the block is μ. What is the minimum force, F, necessary to keep the block at rest?
(A) μmg (B) mgcosθ (C) mgsinθ (D) mgsinθ/μ (E) mg(sinθ – μcosθ)/μ
The normal force on the block can be found from ΣFy = 0 = FN – mgcosθ – F. The force of
friction necessary to hold the block in place is mgsinθ. Setting the force of friction equal to
mgsinθ gives μFN= mgsinθ = μ(F + mgcosθ)
6.
In the figure above, a box moves with speed 5.0 m/s at the bottom of a rough, fixed inclined plane. The box
slides with constant acceleration to the top of the incline as it is being pushed directly to the left with a
constant force of F = 240 N. The box, of mass m = 20.0 kg, has a speed of 2.50 m/s when it reaches the top
of the incline. What is the magnitude of the acceleration of the box as it slides up the incline?
(A) 12.0 m/s2 (B) 10.0 m/s2 (C) 5.88 m/s2 (D) 1.88 m/s2 (E) 0.938 m/s2
7.
A frictionless air puck of mass m is placed on a plane surface inclined at an angle of 60° with respect to the
horizontal. A string of length l is attached to the puck at one end and the upper edge of the inclined plane at
the other to constrain the movement of the puck. What would be the magnitude of the normal force from
the plane acting on the puck?
(A) mg(sin 60°) (B) mg(cos 30°) (C) mg(tan 30°) (D) 𝑚tan 60 (E) None of these
This is really no different than any other incline problem. The normal force on an incline with
no other forces acting into the incline is mgcosθ
Free Response
1. Find the tension in each cable supporting the 600.0 N cat burglar.
37.0
Solution to Problem 1
(a) Since the burglar is held in equilibrium, the tension
in the vertical cable equals the burglar’s weight of
600 N Now, consider the junction in the three
cables:
Fy  0 , giving T2 sin 37.0  600 N  0
Problem 2 Solutions
2. During AP Physics, a student is asked to find the µ of her shoe, in other words, the coefficient of
friction between her shoe and a wooden board provided for her. She places the shoe on the
board and slowly raises one end until the exact moment the shoes starts to slip. Write an
equation for µ in terms of m, θ, and g.
a. Write an equation for µ in terms of m, θ, and g.
𝜇 = 𝑡𝑎𝑛𝜃
b. The student then places a weight in her shoe. What is the µ of the shoe now?
The equation above does not involve the mass, so it will not affect the µ.
c. After doing all this math, the student feels she can confidently and easily find the µ of other
shoes. She walks around the room, looking at the other students’ data. She makes a chart of
everyone’s µ. She puts a heart around the µ of her crush’s shoe, because she’s just that creepy.
Be creepy like her and fill in the chart below.
Shoe
Angle where shoe began to slip
µ of the shoe
Derek’s shoe
Cindy’s shoe
Kenny’s shoe
37
15
22
0.75
0.26
0.40