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WEDGES AND FRICTIONAL FORCES ON FLAT BELTS
Today’s Objectives:
Students will be able to:
a) Determine the forces on a wedge.
b) Determine the tensions in a belt.
APPLICATIONS
Wedges are used to adjust the
elevation or provide stability for
heavy objects such as this large
steel vessel.
How can we determine the
force required to pull the
wedge out?
When there are no applied
forces on the wedge, will it
stay in place (i.e., be selflocking) or will it come out on
its own? Under what physical
conditions will it come out?
APPLICATIONS
(continued)
Belt drives are commonly
used for transmitting power
from one shaft to another.
How can we decide that the
belts will function properly,
i.e., without slipping or
breaking?
APPLICATIONS
(continued)
In the design of a band brake, it
is essential to analyze the
frictional forces acting on the
band (which acts like a belt).
How can we determine the
tensions in the cable pulling on
the band?
How are these tensions, the
applied force P and the torque M,
related?
ANALYSIS OF A WEDGE
W
A wedge is a simple machine in which a
small force P is used to lift a large weight W.
To determine the force required to push the
wedge in or out, it is necessary to draw FBDs
of the wedge and the object on top of it.
It is easier to start with a FBD of the wedge
since you know the direction of its motion.
Note that:
a) the friction forces are always in the
direction opposite to the motion, or impending
motion, of the wedge;
b) the friction forces are along the contacting
surfaces; and,
c) the normal forces are perpendicular to the
contacting surfaces.
ANALYSIS OF A WEDGE (continued)
Next, a FBD of the object on top of the wedge
is drawn. Please note that:
a) at the contacting surfaces between the
wedge and the object the forces are equal in
magnitude and opposite in direction to those
on the wedge; and, b) all other forces acting on
the object should be shown.
To determine the unknowns, we must apply EofE,  Fx = 0 and
 Fy = 0, to the wedge and the object as well as the impending
motion frictional equation, F = S N.
Now of the two FBDs, which one should we start analyzing first?
We should start analyzing the FBD in which the number of
unknowns are less than or equal to the number of equations.
ANALYSIS OF A WEDGE (continued)
W
If the object is to be lowered, then the wedge
needs to be pulled out. If the value of the
force P needed to remove the wedge is
positive, then the wedge is self-locking, i.e.,
it will not come out on its own.
However, if the value of P is negative, or
zero, then the wedge will come out on its
own unless a force is applied to keep the
wedge in place. This can happen if the
coefficient of friction is small or the wedge
angle  is large.
BELT ANALYSIS
Belts are used for transmitting power or
applying brakes. Friction forces play an
important role in determining the
various tensions in the belt. The belt
tension values are then used for
analyzing or designing a belt drive or a
brake system.
BELT ANALYSIS
(continued)
Consider a flat belt passing over a fixed
curved surface with the total angle of
contact equal to  radians.
If the belt slips or is just about to slip,
then T2 must be larger than T1 and the
friction forces. Hence, T2 must be
greater than T1.
Detailed analysis (please refer to your textbook) shows that
T2 = T1 e   where  is the coefficient of static friction
between the belt and the surface. Be sure to use radians when
using this formula!!
EXAMPLE
Given: The load weighs 100 lb and the
S between surfaces AC and BD
is 0.3. Smooth rollers are placed
between wedges A and B.
Assume the rollers and the
wedges have negligible weights.
Find: The force P needed to lift the load.
Plan:
1. Draw a FBD of wedge A. Why do A first?
2. Draw a FBD of wedge B.
3. Apply the E-of-E to wedge B. Why do B first?
4. Apply the E-of-E to wedge A.
EXAMPLE (continued)
N2 10º
The FBDs of wedges A and B are shown in
the figures. Applying the EofE to wedge
B, we get
P
+  FX =
A
N2 sin 10 – N3 = 0
F1= 0.3N1
+  FY = N2 cos 10 – 100 – 0.3 N3 = 0
100 lb
Solving the above two equations, we get
F3= 0.3N3
N2 = 107.2 lb and N3 = 18.6 lb
B
Applying the E-of-E to the wedge A, we get
+  FY = N1 – 107.2 cos 10 = 0;
N1
N2
10º
N1 = 105.6 lb
+  FX = P – 107.2 sin 10 – 0.3 N1 = 0;
P = 50.3 lb
N3
GROUP PROBLEM SOLVING
Given: Blocks A and B weigh 50 lb and
30 lb, respectively.
Find: The smallest weight of cylinder D
which will cause the loss of static
equilibrium.
GROUP PROBLEM SOLVING (continued)
Plan:
1. Consider two cases: a) both blocks slide together, and,
b) block B slides over the block A.
2. For each case, draw a FBD of the block(s).
3. For each case, apply the EofE to find the force needed to
cause sliding.
4. Choose the smaller P value from the two cases.
5. Use belt friction theory to find the weight of block D.