Incline Plane Problems
1) For the ramp below, what is the minimum amount of mass, M, needed to start the block moving
up the ramp? What is the mechanical advantage of the ramp? Assume the ramp and pulley are
frictionless.
500kg
M
30°
2m
2) Repeat problem 1 but with a coefficient of friction of 0.15 between the block and the ramp.
a. What is the actual mechanical advantage?
b. What is the efficiency of the ramp?
3) An incline plane has a slope of 25° and is used to raise a crate to a height of 2 ft and a coefficient
of friction of 0.3. The crate has a an unknown weight and the actual effort to move the box is
300lb. Find:
a. IMA of the ramp
b. AMA of the ramp
c. Weight of the box if.
d. Efficiency of the ramp with a coefficient of friction
4) An incline plane is used to raise a crate that weighs 500lb 4 feet. If the AMA is 3.5.
a. What is applied load to move the crate?
b. What is the length of the slope of the inline plane?
c. What is the angle of the incline plane?
d. What is IMA for this incline plane?
e. What is the friction coefficient for the ramp and box?
f.
5) Wedge A and B are to be glued together. Determine the minimum coefficient of stat friction
Wedges
A andto
B arehold
to be glued
together.
Determinein
the the position shown while the glue
required if clamp CDE is to2.minimum
be
able
the
wedges
coefficient of static friction required, if clamp
CDE is to be able to hold the wedges in the position shown,
dries.
while the glue dries.
A
C
D
10°
10°
B
E
6)
1. If the coefficient of static friction equals 0.3 for all
of contact, determine the smallest value of force P
If the coefficient ofsurfaces
static friction
equals 0.3 for all surfaces of contact, determine the smallest
necessary to raise the block A. Neglect the weight of the
wedge B. to raise the block A. Neglect the weight of the Wedge B.
value of force P necessary
A
300 kg
10°
P
B
A
f AB
2 kip
NAB
7°
P1
B
C
P2
2 Equations of equilibrium for block A
+ Fx = 0: f

AB = 0
(1)
Therefore, f AB = 0
Fy = 0: NAB  2 kip = 0
+

7)
3. Determine the smallest values of forces P1 and P2
1 Free-body diagram of block A
required to raise block A while preventing A from
Determine
the smallest
value ofof forces
P1 and P2 required to raise block A while preventing A
moving horizontally.
The coefficient
static friction
A
for allmoving
surfaces horizontally.
of contact is 0.3, The
and the
weight of of static friction for all surfaces of contact is 0.3 and
from
coefficient
wedges B and C is negligible compared to the weight
the
weight
2 kip
of block
A. of wedges B and C is negligible compared to the weight of block A.
(2)
Solving gives
NAB = 2 kip
3 The friction force f AB has to be zero, since we know that block A is
not to move horizontally and no other horizontal force acts. In fact, we
could have just shown f AB = 0 on the free body initially.
8) To split the log shown a 120 lb force is applied to the top of the wedge, which causes the wedge
to be about to slip further into the log. Determine the friction and normal forces acting on the
sides of the wedge if the coefficient of static friction is 0.6. Also determine if the wedge will pop
out of the log if the force is removed. Neglect the weight of the wedge.
6. To split the log shown, a 120-lb force is applied to the top of the
wedge, which causes the wedge to be about to slip farther into the log.
Determine the friction and normal forces acting on the sides of the
wedge, if the coefficient of static friction is 0.6. Also determine if the
wedge will pop out of the log
if the
120
Lbforce is removed. Neglect the
weight of the wedge.
120 lb
Wedge angle = 8°