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 the block and the plane are ms =0.25 and mk = 0.20. Determine whether the block is in equilibrium, and find the value of the friction force. 1. 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, +240 lb 2. Maximum friction force. N= The F required to maintain equilibrium is directed up and to the right; the tendency of the block is move down the plane. 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 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 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. y x q 1. Assume equilibrium: SFy =N – 250cos30o-50sin30o = 0 N = +241.5 lb q N F SFx =F– 250sin30o +50cos30o = 0, F = +81.7 lb 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 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 y W= 981 N 30o P x 20o F 30o N 1. Determination of F and N: SFy= Psin 20o + N – Wcos 30o = 0, N = 644.36 N SFx= Pcos 20o – F – Wsin 30o = 0, F = 73.32 N 2. Determination of Fmaximum Fm =msN= (0.20)(644.36)=128.87N Since Fm > 73.32, then the block is in equilibrium. y b) Determine the minimum force P to prevent motion 981 N 30o Pmin 20o x 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 30o N 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 c) Determine the maximum force P for which the system is in equilibrium y The maximum force P will be required when motion of the block up the incline is impending. 981 N 30o Pmax 20o x F 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 30o N Solving simultaneously: N = 626 N Pmax = 655N Center of Gravity Additional Examples Centroids – Simple Example for a Composite Body Find the centroid of the given body Centroids – Simple Example for a Composite Body To find the centroid, x 1 AT y 1 AT xA i i y i Ai A2 A1 Determine the area of the components 1 m m 60 m m 3600 m m 2 A2 120 m m 100 m m 12000 m m 2 A1 2 120 Centroids – Simple Example for a Composite Body The total area is x1 b 3 b 40 m m 3 y1 h1 x2 120 m m h1 3 40 m m 3 120 m m 2 y 2 h1 60 m m 60 m m A1 60 m m 2 h2 3 60 m m A2 100 m m 2 110 m m Centroids – Simple Example for a Composite Body To centroid of each component AT A1 A2 3600 m m 12000 m m 2 15600 m m 2 Compute the x centroid x 1 AT x i Ai 1 2 A2 A1 40 m m 3600 m m 2 60 m m 12000 m m 2 2 15600 m m 55.38 m m Centroids – Simple Example for a Composite Body To centroid of each component AT A1 A2 3600 m m 12000 m m 2 15600 m m 2 Compute the y centroid y 1 AT y i Ai 1 2 A2 A1 40 m m 3600 m m 2 110 m m 12000 m m 2 2 15600 m m 93.85 m m Centroids – Simple Example for a Composite Body The problem can be done using a table to represent the composite body. Body Triangle Square Area(mm2) 3600 12000 Sum 15600 centroid (x) centroid (y) x (mm) 40 60 55.38 mm 93.85 mm y(mm) 40 110 x*Area (mm3) y*Area (mm3) 144000 144000 720000 1320000 864000 1464000 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 a large square and subtract the small triangular area. A1 A2 Centroids – Simple Example for a Composite Body The problem can be done using a table to represent the composite body. Body Square Triangle Area(mm2) 19200 -3600 Sum 15600 centroid (x) centroid (y) x (mm) 60 80 55.38 mm 93.85 mm y(mm) 80 20 x*Area (mm3) y*Area (mm3) 1152000 1536000 -288000 -72000 864000 1464000 Centroids –Example for a Composite Body Find the centroid of the given body Centroids –Example for a Composite Body Determine the area of the components A1 1 2 90 m m 60 m m 2700 m m 2 A2 A2 120 m m 90 m m 10800 m m 2 A1 40 m m 2 2513.3 m m 2 A3 2 A3 Centroids –Example for a Composite Body The total area is x1 b 90 m m 3 3 h1 y1 h1 x2 b 30 m m 60 m m 60 m m 3 2 A1 45 m m 2 y 2 h1 h2 x3 b 4r 60 m m 3 3 A2 3 90 m m 40 m m 120 m m 120 m m 2 90 m m 4 40 m m 3 73.02 m m y 3 60 m m 20 m m 40 m m 120 m m A3 Centroids – Example for a Composite Body Body Triangle Square Hemisphere Area(mm2) 2700 10800 -2513.27 Sum 10986.73 centroid (x) centroid (y) x (mm) 30 45 73.02 y(mm) 40 120 120 x*Area (mm3) y*Area (mm3) 81000 108000 486000 1296000 -183528.00 -301592.89 383472.00 34.90 mm 100.34 mm x 1 AT xA i i 383472.00 m m 10986.73 m m 3 2 34.90 m m y 1 AT y i Ai 100.34 m m 1102407.11 m m 10986.73 m m 2 3 1102407.11 Centroids – Example for a Composite Body An Alternative Method would be to subtract to areas Body Triangle Square Hemisphere Area(mm2) -2700 16200 -2513.27 Sum 10986.73 centroid (x) centroid (y) x (mm) 60 45 73.02 34.90 mm 100.34 mm y(mm) 20 90 120 x*Area (mm3) y*Area (mm3) -162000 -54000 729000 1458000 -183528.00 -301592.89 383472.00 1102407.11 Centroids – Class Problem Find the centroid of the body Centroids – Class Problem Find the centroid of the body