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Friction Additional Examples. Example 1: A 100-lb force acts as shown on a 300-lb block placed on an inclined plane. The coefficients of friction between.

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Presentation on theme: "Friction Additional Examples. Example 1: A 100-lb force acts as shown on a 300-lb block placed on an inclined plane. The coefficients of friction between."— Presentation transcript:

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  s =0.25 and  k = Determine whether the block is in equilibrium, and find the value of the friction force.

3 1.Force required for equilibrium: Assuming that F is directed down  Fx = 100lb – 3/5 (300lb) – F = 0 F = 80 lb  Fy = 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. F m =  s N, F m = 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: F actual = F k ( the body is moving) F actual = F k =  k 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  = 30 o and P = 50lb.

6   N F x y 1.Assume equilibrium:  F y =N – 250cos30 o -50sin30 o = 0 N = lb  F x =F– 250sin30 o +50cos30 o = 0, F = lb 2. Maximum friction force: F m =  s N = 0.3 (241.5 lb) = 72.5 lb Since F > F m, then the block moves down Friction force: F =  k N = 0.2(241.5lb) = 48.3 lb

7 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 20 o 30 o P

8 20 o 30 o P x y W= 981 N F N 30 o 1.Determination of F and N:  F y = Psin 20 o + N – Wcos 30 o = 0, N = N  F x = Pcos 20 o – F – Wsin 30 o = 0, F = N 2. Determination of F maximum F m =  s N= (0.20)(644.36)=128.87N Since F m > 73.32, then the block is in equilibrium.

9 20 o 30 o P min x y 981 N F N 30 o b) Determine the minimum force P to prevent motion 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:  F x = P min cos 20 o + F – 981sin 30 o = 0  F x = P min cos 20 o + 0.2N – 981 sin 30 o = 0  F y = P min sin 20 o + N – 981cos 30 o = 0 Then N = 724 N, P min = 368 N

10 c) Determine the maximum force P for which the system is in equilibrium 20 o 30 o P max x y 981 N F N 30 o 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:  F x = P max cos 20 o – 0.2 N – 981 sin 30 o = 0  F y = P max sin 20 o + N – 981 cos 30 o = 0 Solving simultaneously: N = 626 N P max = 655N

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 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


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