Download presentation

1
Friction Additional Examples

2
Example 1: A 100-lb force acts as shown on a 300-lb block placed on an inclined plane. The coefficients of friction between the block and the plane are ms =0.25 and mk = Determine whether the block is in equilibrium, and find the value of the friction force.

3
**Force required for equilibrium: Assuming that F is directed down**

SFx = 100lb – 3/5 (300lb) – F = 0 F = 80 lb SFy = N – 4/5(300lb) = 0, N = +240 lb The F required to maintain equilibrium is directed up and to the right; the tendency of the block is move down the plane. 2. Maximum friction force. Fm = ms N, Fm = 0.25(240 lb) = 60 lb Since the value of F required to maintain equilibrium (80lb) is larger than the maximum possible (60lb), the block will slide down the plane

4
**3. Actual value of friction force:**

Factual = Fk ( the body is moving) Factual = Fk = mk N = 0.2(240lb) = 48 lb The sense of this force is opposite to the sense of motion. The forces acting on the block are not balanced, the resultant is: 3/5 (300lb) – 100lb – 48lb = 32lb

5
Example2: Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when q = 30o and P = 50lb.

6
**y x q q N F Assume equilibrium: SFy =N – 250cos30o-50sin30o = 0**

N = lb SFx =F– 250sin30o +50cos30o = 0, F = lb x q q N F 2. Maximum friction force: Fm = ms N = 0.3 (241.5 lb) = 72.5 lb Since F > Fm, then the block moves down Friction force: F = mk N = 0.2(241.5lb) = 48.3 lb

7
**Example 3: The block in the figure has a mass of 100 kg**

Example 3: The block in the figure has a mass of 100 kg. The coefficient of friction between the block and the inclined surface is 0.2. (a) Determine if the system is in equilibrium when P= 600 N P 20o 30o

8
**W= 981 N y P x 30o 20o F N 30o Determination of F and N:**

SFy= Psin 20o + N – Wcos 30o = 0, N = N SFx= Pcos 20o – F – Wsin 30o = 0, F = N 2. Determination of Fmaximum Fm =msN= (0.20)(644.36)=128.87N Since Fm > 73.32, then the block is in equilibrium. P x 30o 20o F N 30o

9
**b) Determine the minimum force P to prevent motion**

y Pmin The minimum P will be required when motion of the block down the incline is impending. F must resist this motion as shown. Equilibrium exists when: SFx = Pmin cos 20o + F – 981sin 30o = 0 SFx = Pmin cos 20o + 0.2N – 981 sin 30o = 0 SFy = Pmin sin 20o + N – 981cos 30o = 0 Then N = 724 N, P min = 368 N x 30o 20o F N 30o

10
**c) Determine the maximum force P for which the system is in equilibrium**

The maximum force P will be required when motion of the block up the incline is impending. For this condition, F will tend to resist this motion as shown. Then: SFx = Pmax cos 20o – 0.2 N – 981 sin 30o = 0 SFy = Pmax sin 20o + N – 981 cos 30o = 0 Solving simultaneously: N = 626 N Pmax = 655N 981 N y Pmax x 30o 20o F N 30o

11
Center of Gravity Additional Examples

12
**Centroids – Simple Example for a Composite Body**

Find the centroid of the given body

13
**Centroids – Simple Example for a Composite Body**

To find the centroid, Determine the area of the components

14
**Centroids – Simple Example for a Composite Body**

The total area is

15
**Centroids – Simple Example for a Composite Body**

To centroid of each component Compute the x centroid

16
**Centroids – Simple Example for a Composite Body**

To centroid of each component Compute the y centroid

17
**Centroids – Simple Example for a Composite Body**

The problem can be done using a table to represent the composite body.

18
**Centroids – Simple Example for a Composite Body**

An alternative method of computing the centroid is to subtract areas from a total area. Assume that area is Assume that area is a large square and subtract the small triangular area.

19
**Centroids – Simple Example for a Composite Body**

The problem can be done using a table to represent the composite body.

20
**Centroids –Example for a Composite Body**

Find the centroid of the given body

21
**Centroids –Example for a Composite Body**

Determine the area of the components

22
**Centroids –Example for a Composite Body**

The total area is

23
**Centroids – Example for a Composite Body**

24
**Centroids – Example for a Composite Body**

An Alternative Method would be to subtract to areas

25
**Centroids – Class Problem**

Find the centroid of the body

26
**Centroids – Class Problem**

Find the centroid of the body

Similar presentations

OK

Chapter 8 - Friction Sections 8.1 - 8.4. Friction w Frictional forces resist movement of a rigid body over a rough surface. w It is assumed that the.

Chapter 8 - Friction Sections 8.1 - 8.4. Friction w Frictional forces resist movement of a rigid body over a rough surface. w It is assumed that the.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google

Ppt on thermal conductivity of insulating powder coating Ppt on data handling for class 1 Download free ppt on p-block elements Ppt on growing old gracefully Ppt on landing gear systems Ppt on complex numbers class 11th result Ppt on hcl company profile Ppt on water resources download Ppt on mobile apps Ppt on conservation of forest and wildlife in india