EGR 181 Fall Quarter 1997Homework SolutionsProblems from Engineering Mechanics - Statics, Meriam and Kraige 4th EditionChapter 3 Problem 1 Page 126The 50-kg homogenous smooth sphere restson the 30o incline and bears against thesmooth vertical wall B. Calculate the contactforces at A and B.1. Mechanical System = Sphere. The sphere isthe appropriate mechanical system because theonly force you know (weight) acts on thesphere and the two forces you want to know(contact forces) also act on the sphere.2. Free Body Diagram = As smooth walls provide noresistance to movement along the wall, the only forcethey can exert is normal or perpendicular to the wall. Thus the free body diagram contains three forces, theweight of the ball and two normal contact forces at Aand B.3. Equations = Sum of Forces = Weight (downward) +Contact at B(horizontal) + Contact at A (diagonallyupward). Note that as the directions of the forces at Aand B are known but their magnitudes are unknown,we express these forces in terms of symbolsrepresenting their magnitude multiplied by unitvectors representing their direction.S F = -50 kg * 9.81 m/s2 j - B i + A ( cos 60 i + sin 60j) = ( 1/2 A - B) i + ( 31/2/2 A - 490.5 N) j = 0 = 0 i + 0 j4. Solve the Equations. For two vectors to be equal, each of their components must be equal. Thuseach of the two components of the resultant force vector must be zero. Noting that the Y componentinvolves only the unknown A, we can readily solve for A.EGR 181 Homework Solutions file://localhost/Users/ØDocuments/08_Fall_2110/C5/C5_Probl...1 of 2 10/5/08 11:56 AMA = 2 * 490.5 N / 31/2 = 566.4 NNow that A is known, B is the only unknown in the X component of the sum of the forces.B = 1/2 A = 283.2 N EGR 181 Homework Solutions file://localhost/Users/ØDocuments/08_Fall_2110/C5/C5_Probl...2 of 2 10/5/08 11:56 AMEGR 181 Fall Quarter 1997Homework SolutionsProblems from Engineering Mechanics - Statics, Meriam and Kraige 4th EditionChapter 3 Problem 13 Page 129Calculate the magnitude F of the forcesupported by the pin at A for the loadedbracket. Neglect the weight of thebracket.1. Mechanical System = Bracket. Thebracket is the appropriate mechanicalsystem as the only force we know (900 N)acts on the bracket and the forces we wantto know (at pin A) act on the bracket. 2. Free Body Diagram. 900 N forceapplied diagonally downward. Weight ofthe bracket is neglected so no weight forceis shown. Contact at B with a smoothroller. The only resistance is to horizontalmovement, so we have only a horizontalforce. The pin at A resists both horizontaland vertical movement so we have both ahorizontal and a vertical component offorce at A. Pin A does permit the bracketto rotate or swing freely so there is nocouple at that point. Note that we havedepicted the forces exerted by the pin A onthe bracket. By the law of action andreaction these forces are equal andopposite to the forces exerted by thebracket on the pin. Thus in finding themagnitude of the force exerted by the pinEGR 181 Homework Solutions file:///Users/ØDocuments/08_Fall_2110/C5/C5_Problems_equil...1 of 3 10/5/08 2:32 PMon the bracket, we are also finding the magnitude of the force exerted on the pin by the bracket.3. Sum of the Forces = 900 N diagonally downward. Horizontal force of unknown magnitude at B. Force at A with unknown horizontal and vertical components. Note that all unknown values havebeen represented by appropriate symbols in the free body diagram.S F = 900 N ( -sin(15) i - cos(15) j ) + B i + AX i + AY j = ( B + AX - 232.9 N ) i + ( AY - 869.3 N) j = 0 = 0 i + 0 j Noting that the force equation contains three unknown quantities but only two components indicatesthat an additional equation will be required. Thus we must also evaluate the sum of the moments. Wecan choose to sum moments about any point. We normally choose a point that will make the momentevaluation as simple as possible. If we sum moments about the pin at A, then the position vector tothe pin forces at A from that point will be zero, and their contribution to the moment will be zero. This appears to be a good choice of points. The position vector of the force at B from point A can beseen to be 0.23 m straight down. The position vector from point A to the 900 N force can be seen tobe 0.23 m straight down and 0.18 m to the right.S MA = 0 X ( AX i + AY j ) + -0.23m j X B i + (-0.23m j + 0.18 m i) X (900 N ( -sin(15) i - cos(15) j) )Recalling that i X i is zero, i X j is k, j X i is -k and j X j is zero, and performing the required crossproducts yields:S MA = 0.23 m B k - 0.23 m 900 N sin(15) k - 0.18 m 900 N cos(15) k = ( 0.23 m B - 53.6 N m - 156.5 N m ) k = 0.23 m B - 210.1 N m ) k = 0 = 0 k4. Solving the Equations. If two vectors are equal, their components must be equal. Thus the X andY components of the resultant force vector must be zero. The Z component of the resultant momentabout point A must be zero. Setting the Y component of the resultant force vector to zero determinesthe Y component of the pin force, AY.AY = 869. 3 NSetting the Z component of the moment equation to zero, determines the force exerted on the bracketat B, B.B = 210.1 N m / 0.23 m = 913.3 NSetting the X component of the resultant force to zero and using the value of the force B justdetermined, yields:AX = 232.9 N - B = -680.3 NEGR 181 Homework Solutions file:///Users/ØDocuments/08_Fall_2110/C5/C5_Problems_equil...2 of 3 10/5/08 2:32 PMThe fact that the result for AX is negative means that the force exerted on the bracket by the pin in thehorizontal direction is in the opposite direction from that shown on the free body diagram. Thus theforce exerted on the bracket by the pin is vertically upward (as shown) and horizontally to the left(opposite of shown). The total force exerted by the pin on the bracket is :A = AX i + AY j = -680.3 N i + 869.3 N jThe magnitude of this force, which is also the magnitude of the force exerted by the bracket on thepin, is given by the square root of the sum of the squares of the components.|| A || = ( 680.32 + 869.32)1/2 N = 1103.9 N EGR 181 Homework Solutions file:///Users/ØDocuments/08_Fall_2110/C5/C5_Problems_equil...3 of 3 10/5/08 2:32 PMEGR 181 Fall Quarter 1997Homework SolutionsProblems from Engineering Mechanics - Statics, Meriam and Kraige 4th EditionChapter 3 Problem 15 Page 129The force P on the handle of the positioning leverproduces a vertical compression of 60 lb in the coiledspring at