Explanation Given information: Set P = 800 N . Assumption: The self-weight of the cable is ignored. The cable is perfectly flexible and inextensible. The tensile force acting in the cable is always tangent to the cable at points along its length. Being inextensible, the cable has a constant length both before and after the load is applied. The cable or a segment of the cable can be treated as a rigid body. Method of joints is used to determine the force in the cable. Consider the state of member as tension where the force is pulling the member and as compression where the force is pushing the member. Consider the force indicating right side as positive and left side as negative in the horizontal components of forces. Consider the force indicating upward is taken as positive and downward as negative in the vertical components of forces. Consider clockwise moment as negative and anticlockwise moment as positive wherever applicable. Apply the Equation of equilibrium wherever applicable. Cable segment ABC : Show the free-body diagram of cable segment ABC as in Figure (1). Using Figure (1), Determine the value of θ : tan θ = Opposite side Adjacent side (I) Moment about point A : Determine the force in cable segment CD by taking moment about point A . ∑ M A = 0 P ( 4 ) − F C D cos θ 1 ( 10 ) − F C D sin θ 1 ( 2 ) + 600 ( 10 ) = 0 (II) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y − F C D sin θ 1 = 0 (III) Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of forces. ∑ F x = 0 − A x − F C D cos θ 1 + P + 600 = 0 (IV) Conclusion: Substitute 1 m for the opposite side and 2 m for the adjacent side in Equation (I). tan θ 1 = 1 2 θ 1 = 26.57 ° Substitute 800 N for P and 26.57 ° for θ 1 in Equation (II). 3 , 200 − F C D cos 26.57 ° ( 10 ) − F C D sin 26.57 ° ( 2 ) + 6 , 000 = 0 9 , 200 − 9.838 F C D = 0 9.838 F C D = 9 , 200 F C D = 935.11 N Substitute 935.11 N for F C D and 26.57 ° for θ 1 in Equation (III). A y − 935.11 sin 26.57 ° = 0 A y = 418.27 N Substitute 935.11 N for F C D , 26.57 ° for θ 1 , and 800 N for P in Equation (IV). − A x − 935.11 cos 26.57 ° + 800 + 600 = 0 − A x − 836.35 + 1 , 400 = 0 A x = 563.65 N Joint A : Show the free-body diagram of the joint A as in Figure (2).

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ISBN: 9780134867267

Author: HIBBELER

Publisher: PEARSON CO

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Chapter 7, Problem 97P

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Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',... S3L,V3x,'k',S3L,V3r,'r',S3L,W3r,'b',U3x,U3y,'g',...... 'LineWidth',2,'MarkerSize',10),... axis([-1 6 -4 4]), ... title('Velocity Triangle'), ... xlabel('x'),ylabel('y'), grid

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The distance x B the force at B acts from A .

Solution Summary: The author explains the method of joints used to determine the force in cable segment ABC by taking moment about point A.

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