1. 8.2 Problems Involving Dry Friction
Example 8.6
Determine the normal force P that must be exerted
on the rack to begin pushing the 100kg pipe up the
20° incline. The coefficients of
static friction at points of
contact are (μs)A = 0.15
and (μs)B = 0.4.

2. 8.2 Problems Involving Dry Friction
View Free Body Diagram
Solution
The rack must exert a force P on the pipe due to force equilibrium in the x direction
4 unknowns
3 equilibrium equations and 1 frictional equation which apply at either A or B
If slipping begins to occur at B, the pipe will roll up the incline
If the slipping occurs at A, the pipe will begin to slide up the incline

3. 8.2 Problems Involving Dry Friction
Solution
FBD of the rack

4. 8.2 Problems Involving Dry Friction
Solution
Pipe rolls up incline

5. 8.2 Problems Involving Dry Friction
Solution
Inequality does not apply and slipping occurs at A
Pipe slides up incline
Solving
Check: no slipping occur at B

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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.

12. 8.3 Wedges
View Free Body Diagram
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

13. 8.3 Wedges
Solution
5 unknowns, 3 equilibrium equations for the stone and 2 for the wedge

14. 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