1. Applications of Friction in Machines: Wedges
A wedge is the simplest of machines and requires friction to function. A wedge is an inclined plane that is used to raise or lower an object.
The diagram to the right shows a simple situation in which a wedge is lifting a block of mass m. This can be analyzed by examining the wedge and block separately, using standard equilibrium techniques or by analyzing vector diagrams.
Looking at each object separately, we see that there are two forces, R1 and R2, which correspond to the resultant of a frictional force and a normal force. These forces will be at an angle f (𝑡𝑎𝑛𝜑=𝜇) relative to a reference line that is normal to the surface. Notice that the reference line on top of the wedge is at an angle a with respect to the vertical. Therefore the angle of R2 with respect to the vertical is f + a.
Vector Diagrams:

2. Similarly it is possible to examine the case where a block is lowered
Similarly it is possible to examine the case where a block is lowered. The applied force and the direction of the friction would be reversed in this case.
The diagram on the right shows the forces that need to be considered. The analysis would be identical to lifting the block otherwise.
Notice that this time the angle between R2 and the vertical is now f – a.
The vector diagram for this situation is shown on the far right.
The diagram at the lower right shows the conditions for self locking. This can occur when P is removed.
The reactions R1 and R2 will be colinear when P is removed and this will create a range of angles where the wedge will not slip.
Self-locking will occur when a < 2f.

3. Applications of Friction in Machines: Screws
A screw is essentially an inclined plane wrapped around a cylinder. Therefore we will use a similar analysis as we did for wedges.
Types of Screws:
Square thread – used to transmit power
V-Thread – used for fastening
The screw has several additional quantities that must be defined.
r – The radius of the thread, measured from the center of the shaft to the center of the thread.
L – This is the lead or the vertical height change (advancement) per revolution.
a – This is the pitch or angle of the thread.
A downward force W and moment M are required for the screw to operate properly. This is true for raising or lowering the screw.

4. To analyze this situation we need to examine both the vertical forces and the moments. Let us begin with the moments.
𝑀 = 𝑟 × 𝐹
→𝑀=𝑟 𝑅𝑠𝑖𝑛𝜃
=𝑅𝑟𝑠𝑖𝑛 𝛼+𝜑
For a single contact point.
The net moment can then be defined as:
𝑀= 𝑀 = 𝑅𝑟𝑠𝑖𝑛 𝛼+𝜑
=𝑟𝑠𝑖𝑛 𝛼+𝜑 𝑅
Now examining the vertical forces:
𝐹 𝑦 = 𝑅𝑐𝑜𝑠 𝛼+𝜑 −W=0
→ 𝑊=𝑐𝑜𝑠 𝛼+𝜑 𝑅
We can now combine the two expressions since they both contain 𝑅 :
𝑊=𝑐𝑜𝑠 𝛼+𝜑 𝑅
=𝑐𝑜𝑠 𝛼+𝜑 𝑀 𝑟𝑠𝑖𝑛 𝛼+𝜑
→𝑀=𝑊𝑟𝑡𝑎𝑛 𝛼+𝜑
In order to use this expression we need to know a. We can determine a by unwrapping one revolution of the thread.
𝑡𝑎𝑛𝛼= 𝐿 2𝜋𝑟
→ 𝛼= 𝑡𝑎𝑛 −1 𝐿 2𝜋𝑟

5. We can similarly analyze the situation by starting with the unwrapped thread.
We begin be defining P as the force required to push the moveable thread up the fixed incline, where 𝑃= 𝑀 𝑟 .
We can now analyze the forces in the x and y directions to determine W.
𝐹 𝑦 =𝑊−𝑅𝑐𝑜𝑠 𝛼+𝜑 =0
→𝑅= 𝑊 𝑐𝑜𝑠 𝛼+𝜑
𝐹 𝑥 =𝑃−𝑅𝑠𝑖𝑛 𝛼+𝜑 =0
→𝑃= 𝑊 𝑐𝑜𝑠 𝛼+𝜑 𝑠𝑖𝑛 𝛼+𝜑
→𝑃=𝑅𝑠𝑖𝑛 𝛼+𝜑
→𝑃= 𝑀 𝑟 =𝑊𝑡𝑎𝑛 𝛼+𝜑
→𝑀=𝑊𝑟𝑡𝑎𝑛 𝛼+𝜑
Note that the self-locking condition for the screw is a < f, since there is only friction on one side of the incline. When a = f the screw is on the verge of unwinding.