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WEDGES AND FRICTIONAL FORCES ON FLAT BELTSPowerPoint Presentation

WEDGES AND FRICTIONAL FORCES ON FLAT BELTS

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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.

- In-Class Activities:
- Check Homework, if any
- Reading Quiz
- Applications
- Analysis of a Wedge
- Analysis of a Belt
- Concept Quiz
- Group Problem Solving
- Attention Quiz

1. A wedge allows a ______ force P to lift

a _________ weight W.

A) (large, large) B) (small, small)

C) (small, large) D) (large, small)

W

2. Considering friction forces and the indicated motion of the belt, how are belt tensions T1 and T2 related?

A) T1 > T2 B) T1 = T2

C) T1 < T2 D) T1 = T2 e

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 self-locking) 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?

A wedge is a simple machine in which a small force P is used to lift a large weight W.

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 areequal 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 whichthe number of unknowns are less than or equal to the number of equations.

ANALYSIS OF A WEDGE (continued)

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.

W

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.

- Wedges - simple machines used to raise heavy loads.
- Force required to lift block is significantly less than block weight.
- Friction prevents wedge from sliding out.
- Want to find minimum force P to raise block.

- Block as free-body

or

or

Wedges- Moment of force Q is equal to moment of force P.

- Self-locking, solve for Q to lower load.

- Impending motion upwards. Solve for Q.

- Non-locking, solve for Q to hold load.

- Square-threaded screws frequently used in jacks, presses, etc. Analysis similar to block on inclined plane. Recall friction force does not depend on area of contact.

- Thread of base has been “unwrapped” and shown as straight line. Slope is 2pr horizontally and lead L vertically.

C-Clamp

A clamp is used to hold two pieces of wood together as shown. The clamp has a double square thread of mean diameter equal to 10 mm with a pitch of 2 mm. The coefficient of friction between threads is ms = 0.30.

If a maximum torque of 40 N*m is applied in tightening the clamp, determine (a) the force exerted on the pieces of wood, and (b) the torque required to loosen the clamp.

- SOLUTION
- Calculate lead angle and pitch angle.

- Using block and plane analogy with impending motion up the plane, calculate the clamping force with a force triangle.

- With impending motion down the plane, calculate the force and torque required to loosen the clamp.

- SOLUTION
- Calculate lead angle and pitch angle. For the double threaded screw, the lead L is equal to twice the pitch.

- Using block and plane analogy with impending motion up the plane, calculate clamping force with force triangle.

- With impending motion down the plane, calculate the force and torque required to loosen the clamp.

BELT ANALYSIS and torque required to loosen the clamp.

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 and torque required to loosen the clamp.(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 and torque required to loosen the clamp.

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.

100 lb and torque required to loosen the clamp.

F3= 0.3N3

B

N3

10º

N2

EXAMPLE (continued)

10º

N2

The FBDs of wedges A and B are shown in the figures. Applying the EofE to wedge B, we get

A

P

- + FX = N2 sin 10 – N3 = 0
- + FY = N2 cos 10 – 100 – 0.3 N3 = 0
Solving the above two equations, we get

N2 = 107.2 lb and N3 = 18.6 lb

F1= 0.3N1

N1

Applying the E-of-E to the wedge A, we get

+ FY = N1– 107.2 cos 10 = 0; N1 = 105.6 lb

+ FX = P – 107.2 sin 10 – 0.3 N1 = 0; P = 50.3 lb

1. Determine the and torque required to loosen the clamp.direction of the friction force on object Bat the contact point between A and B.

A) B)

C) D)

CONCEPT QUIZ

2. The boy (hanging) in the picture weighs 100 lb and the woman weighs 150 lb. The coefficient of static friction between her shoes and the ground is 0.6. The boy will ______ ?

A) be lifted up. B) slide down.

C) not be lifted up. D) not slide down.

GROUP PROBLEM SOLVING and torque required to loosen the clamp.

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 and torque required to loosen the clamp.(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.

GROUP PROBLEM SOLVING and torque required to loosen the clamp.(continued)

- Case a:
- + FY = N – 80 = 0
N = 80 lb

- + FX = 0.4 (80) – P = 0
P = 32 lb

P

30 lb

B

50 lb

A

F=0.4 N

N

30 lb and torque required to loosen the clamp.

P

0.6 N

20º

N

GROUP PROBLEM SOLVING (continued)

Case b:

+ Fy = N cos 20 + 0.6 N sin 20 – 30 = 0 N = 26.20 lb

+ Fx = – P + 0.6 ( 26.2 ) cos 20 – 26.2 sin 20 = 0 P = 5.812 lb

Case b has the lowest P and will occur first. Next, using the frictional force analysis of belt, we get

WD = P e = 5.812 e 0.5 ( 0.5 ) = 12.7 lb

A Block D weighing 12.7 lb will cause the block B to slide over the block A.

ATTENTION QUIZ and torque required to loosen the clamp.

1. When determining the force P needed to lift the block of weight W, it is easier to draw a FBD of ______ first.

A) the wedge B) the block

C) the horizontal ground D) the vertical wall

W

2. In the analysis of frictional forces on a flat belt, T2 = T1 e . In this equation, equals ______ .

A) angle of contact in degrees B) angle of contact in radians

C) coefficient of static friction D) coefficient of kinetic friction

End of the Lecture and torque required to loosen the clamp.

Let Learning Continue

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