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 = 0F = 80 lbSFy = N – 4/5(300lb) = 0, N = +240 lbThe 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 lbSince 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 lbThe 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 = lbSFx =F– 250sin30o +50cos30o = 0, F = lbxqqNF2. Maximum friction force:Fm = ms N = 0.3 (241.5 lb) = 72.5 lbSince F > Fm, then the block moves downFriction 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 NP20o30o
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 = NSFx= Pcos 20o – F – Wsin 30o = 0, F = N2. Determination of FmaximumFm =msN= (0.20)(644.36)=128.87NSince Fm > 73.32, thenthe block is in equilibrium.Px30o20oFN30o
9 b) Determine the minimum force P to prevent motion yPminThe 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 = 0SFx = Pmin cos 20o + 0.2N – 981 sin 30o = 0SFy = Pmin sin 20o + N – 981cos 30o = 0Then N = 724 N, P min = 368 Nx30o20oFN30o
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 = 0SFy = Pmax sin 20o + N – 981 cos 30o = 0Solving simultaneously:N = 626 NPmax = 655N981 NyPmaxx30o20oFN30o
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 componentCompute the x centroid
16 Centroids – Simple Example for a Composite Body To centroid of each componentCompute 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 isAssume 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
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