Explanation Given information: The dimension of the raft is 5 × 5 m . The weight of the children A ( F A ) is 375 N. The weight of the children B ( F B ) is 260 N. The weight of the children C ( F C ) is 400 N. The weight of the fourth child ( F D ) is 425 N. Calculation: Show the raft with the children as in Figure (1). Show the free body diagram of the raft as in Figure (2). Refer to Figure (2). The point of application of the resultant from x axis as x D (from OG ) and from z axis as z D (from OE ). Calculate the resultant forces by resolving the force equilibrium: ∑ F = R − F A − F B − F C − F D = − R Substitute 375 N for F A , 260 N for F B , 400 N for F C , and 425 N for F D . − 375 j − 260 j − 400 j − 425 j = − R R = ( 1460 N ) j Calculate the distance ( z A ) of the children A standing from point O (from y axis): z A = 5 − 2 m =3 m Calculate the distance ( z B ) of the children B standing from point O (from y axis): z B = 0.5 m Calculate the distance ( z C ) of the children C standing from point O (from y axis): z C = 5 − 0.25 m =4 .75 m Calculate the distance ( z H ) of the resultant force acting on the raft: Since the resultant force is acting in the center of the raft. z H = 5 2 =2 .5 m Calculate the point of application of the resultant of the four weights: Take the moment about x axis. ∑ M x = 0 F A ( z A ) + F B ( z B ) + F C ( z C ) + F D ( z D ) − R ( z H ) = 0 Substitute 375 N for F A , 260 N for F B , 400 N for F C , 425 N for F D , 3 m for z A , 0

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ISBN: 9781259633133

Author: BEER

Publisher: MCG

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Chapter 3.4, Problem 3.128P

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The line of action of the resultant of the four weights is to pass through the center of the raft.

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Chapter 3.4, Problem 3.127P

Chapter 3.4, Problem 3.129P

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The answer should equal to 1157. Please sent me the solution. Thank you!

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