(a) Explanation Given information: The speed of the car B ( v B ) is 45 mi/hr. The speed of the car C ( v C ) is 60 mi/hr. The weight of the car A ( W A ) is 3,000 lb. The weight of the car B ( W B ) is 2,600 lb. The weight of the car C ( W C ) is 2,400 lb. The co-ordinates of the utility pole ( x P , y P ) is ( 46 ft , 59 ft ) . Calculation: Draw the free body diagram of the system as shown in Figure (1). The origin is set at the point O where the cars A and B collide. The position of the initial mass center for the first collision is found out. At the origin, the value of x and y is zero. From the definition of mass center: ( W A + W B + W C ) ( x ¯ 0 i + y ¯ 0 j ) = W A ( x A i + y A j ) + W B ( x B i + y B j ) + W C ( x C i + y C j ) Here, position of center of mass in x direction is x ¯ 0 , position of center of mass in y direction is y ¯ 0 , position of C in the x direction is x C , and position of C in the y direction is y C . Substitute 3,000 lb for W A , 2,600 lb for W B , 2,400 lb for W C , 0 for x A , 0 for x B , 0 for y A , 0 for y B , 32 ft for x C , and 10 ft for y C . [ ( 3 , 000 lb + 2 , 600 lb + 2 , 400 lb ) × ( x ¯ 0 i + y ¯ 0 j ) ] = [ ( 3 , 000 lb ) ( 0 i + 0 j ) + ( 2 , 600 lb ) ( 0 i + 0 j ) + ( 2 , 400 lb ) ( ( 32 ft ) i + ( 10 ft ) j ) ] 8 , 000 ( x ¯ 0 i + y ¯ 0 j ) = 76 , 800 i + 24 , 000 j 8 , 000 x ¯ 0 i + 8 , 000 y ¯ 0 j = 76 , 800 i + 24 , 000 j (1) Find the position of center of mass in x direction x ¯ 0 : Equate i vectors from Equation (1). 8 , 000 x ¯ 0 i = 76 , 800 i x ¯ 0 = 76 , 800 i 8 , 000 i x ¯ 0 = 9.6 ft Find the position of center of mass in y direction y ¯ 0 : Equate j vectors from Equation (1) (b) To determine The speed of the car A ( v A ) .

11th Edition

ISBN: 9781259633133

Author: BEER

Publisher: MCG

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Chapter 14.1, Problem 14.20P

(a)

To determine

The time elapsed from the first collision to the stop at P.

(b)

To determine

The speed of the car A (vA).

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Chapter 14.1, Problem 14.19P

Chapter 14.1, Problem 14.21P

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