1. Answers to Inclined Plane Problems
1. A box of mass 25 kg is placed on an inclined plane that is at a
40o angle with the horizontal.
m = 25 kg
(a) Draw a sketch.
405o
Fperp
40o
Wt.
Fpar

2. 1. A box of mass 25 kg is placed on an inclined plane that is at a
40o angle with the horizontal.
(b) Determine the weight of the box.
m = 25 kg
Wt. = m g
= ( 25 kg )( 9.8 m/s2 )
40o
Fperp
40o
Wt.
Wt. = 245 N
Fpar

3. 1. A box of mass 25 kg is placed on an inclined plane that is at a
40o angle with the horizontal.
(c) Determine the normal force FN and the
parallel force Fp.
m = 25 kg
FN = Fperp
40o
Fperp
40o
245 N
Fpar
Fpar
Fperp
sin 40 =
cos 40 =
245
245
Fpar = 245 sin 40
Fperp = 245 cos 40
Fperp = 188 N
Fpar = 157 N
FN = 188 N

4. 1. A box of mass 10 kg is placed on an inclined plane that is at a
25o angle with the horizontal.
(d) Assume the plane is frictionless. Find
the acceleration of the
block down the plane.
m = 25 kg
a = ?
40o
188 N
40o
F = m a
245 N F
F = Fpar
157 N
a = m
157 N =
25 kg
a = 6.3 m/s2

5. 2. A 1500-kg car is on an inclined slope at an angle of 20o with the
horizontal. It has been left in neutral without the parking brake on.
(a) Determine FN and Fp.
m = 1500 kg
Wt. = m g
= ( 1500 kg )( 9.8 m/s2 )
Fperp
20o
20o
Wt. = N N
Fpar
Fperp
Fpar
cos 20 =
sin 20 =
14 700
14 700
Fperp = cos 20
Fpar = sin 20
Fperp = N
Fpar = N
FN = N

6. 2. A 1500-kg car is on an inclined slope at an angle of 20o with the
horizontal. It has been left in neutral without the parking brake on.
(b) If the coefficient of friction is 0.35,
calculate the force of friction.
m = 1500 kg
Ff = μ FN N
20o
20o N
= ( 0.35 )( N )
5028 N
Ff = N

7. 2. A 1500-kg car is on an inclined slope at an angle of 20o with the
horizontal. It has been left in neutral without the parking brake on.
(c) Will the car begin to slide
down the slope?
m = 1000 kg
4834 N N
20o
20o N
5028 N
Fp > Ff , so car begins to slide
down the slope

8. 3. A crate of mass 80 kg is placed on an inclined plane at an angle of
40o with the horizontal. The coefficient of friction between the crate
and the plane is Find the acceleration of the crate down the plane.
Wt. = m g
m = 80 kg
= ( 80 kg )( 9.8 m/s2 )
a = ?
Wt. = 784 N
Fperp
40o
40o
Wt.
Fpar

9. 3. A crate of mass 80 kg is placed on an inclined plane at an angle of
40o with the horizontal. The coefficient of friction between the crate
and the plane is Find the acceleration of the crate down the plane.
Wt. = m g
m = 80 kg
= ( 80 kg )( 9.8 m/s2 )
a = ?
Wt. = 784 N
Fperp
40o
40o
784 N
Fperp
cos 40 =
Fpar
Fpar
784
sin 40 =
784
Fperp = 784 cos 40
Fpar = 784 sin 40
Fperp = 601 N
Fpar = 504 N
FN = 601 N

10. 3. A crate of mass 80 kg is placed on an inclined plane at an angle of
40o with the horizontal. The coefficient of friction between the crate
and the plane is Find the acceleration of the crate down the plane.
m = 80 kg Ff
Ff = μ FN
a = ?
= ( 0.20 )( 601 N )
601 N
Ff = 120 N
40o
40o
784 N
504 N

11. 3. A crate of mass 80 kg is placed on an inclined plane at an angle of
40o with the horizontal. The coefficient of friction between the crate
and the plane is Find the acceleration of the crate down the plane.
m = 80 kg
120 N
Fnet = Fpar + Ff
a = ?
= ( +504 N ) + ( N )
Fnet
601 N
40o
Fnet = 384 N
40o
784 N F
F = m a
a =
504 N m
384 N =
80 kg
a = 4.8 m/s2

12. 4. A man wishes to push a refrigerator up a ramp that is at an angle
of 15o with the ground. The refrigerator weighs 1000 N. The
coefficient of friction is 0.14.
(a) Determine the components FN and Fp.
15o
Fperp
15o
1000 N
Fperp
Fpar
Fpar
cos 15 =
sin 15 =
1000
1000
Fperp = cos 15
Fpar = sin 15
Fperp = 966 N
Fpar = 259 N
FN = 966 N

13. 4. A man wishes to push a refrigerator up a ramp that is at an angle
of 15o with the ground. The refrigerator weighs 1000 N. The
coefficient of friction is 0.14.
(a) Determine the components FN and Fp.
(b) Determine the force of friction.
15o
966 N
Ff = μ FN
15o
1000 N
= ( 0.14 )( 966 N )
259 N
Ff = 135 N

14. 4. A man wishes to push a refrigerator up a ramp that is at an angle
of 15o with the ground. The refrigerator weighs 1000 N. The
coefficient of friction is 0.14.
(a) Determine the components FN and Fp.
(b) Determine the force of friction.
(c) What force must the man exert to
push the safe up the ramp?
FA = ?
135 N
15o
966 N
15o
1000 N
Man must exert enough force to
overcome both Fpar and Ff
259 N
so FA = ( 259 N ) + ( 135 N )
FA = 394 N