1. 8.3 Wedges
A simple machine used to transform an applied force
into much larger forces, directed at approximately
right angles to the applied force
Used to give small displacements or adjustments to
heavy load
Consider the wedge used to
lift a block of weight W by
applying a force P to the wedge

2. 8.3 Wedges
FBD of the block and the wedge
Exclude the weight of the wedge since it is small
compared to weight of the block

3. 8.3 Wedges
Frictional forces F1 and F2 must oppose the
motion of the wedge
Frictional force F3 of the wall on the block must
act downward as to oppose the block’s upward
motion
Location of the resultant forces are not important
since neither the block or the wedge will tip
Moment equilibrium equations not considered
7 unknowns - 6 normal and frictional force and
force P

4. 8.3 Wedges
2 force equilibrium equations (∑Fx = 0, ∑Fy = 0)
applied to the wedge and block (4 equations in
total) and the frictional equation (F = µN)
applied at each surface of the contact (3
equations in total)
If the block is lowered, the frictional forces will
act in a sense opposite to that shown
Applied force P will act to the right if the
coefficient of friction is small or the wedge angle
θ is large

5. 8.3 Wedges
Otherwise, P may have the reverse
sense of direction in order to pull the
wedge to remove it
If P is not applied or P = 0, and friction
forces hold the block in place, then the
wedge is referred to as self-locking

6. 8.3 Wedges
Example 8.7
The uniform stone has a mass of 500kg and is held
in place in the horizontal position using a wedge at
B. if the coefficient of static friction µs = 0.3, at the
surfaces of contact, determine the minimum force
P needed to remove the wedge. Is the wedge self-
locking? Assume that the stone does not slip at A.

7. 8.3 Wedges
Solution
Minimum force P requires F = µs NA at the
surfaces of contact with the wedge
FBD of the stone and the wedge
On the wedge, friction force opposes the motion
and on the stone at A, FA ≤ µsNA, slipping does
not occur

8. 8.3 Wedges
Solution
5 unknowns, 3 equilibrium equations for the
stone and 2 for the wedge
∑ M A = 0;
− 4905 N (0.5m) + ( N B cos 7 o N )(1m) + (0.3N B sin 7 o N )(1m) = 0
+ → ∑ Fx = 0;
2383.1sin 7 o − 0.3(2383.1 cos 7 o ) + P − 0.3 N C = 0
+ ↑ ∑ Fy = 0;
N C − 2383.1 cos 7 o N − 0.3(2383.1sin 7 o ) = 0
N C = 2452.5 N , P = 1154.9 N = 1.15kN

9. 8.3 Wedges
Solution
Since P is positive, the wedge must be pulled
out
If P is zero, the wedge would remain in place
(self-locking) and the frictional forces
developed at B and C would satisfy
FB < µsNB
FC < µsNC