Explanation Given information: The weight of the empty car is m f = 0.4 lb . The velocity of the car is u = 6 ft / s . The weight of water is m = 2 lb . Calculation: Consider the mass of the car plus the water in the tank as m at the beginning of the interval and the mass at the end of the interval as m − Δ m . Calculate the weight of the car with water m 0 as shown below. m 0 = 0.4 + 2 = 2.4 lb Consider the velocity of the car as v . Sketch the Free Body Diagram of the car as shown in Figure 1. Refer to Figure 1. Apply the impulse and momentum principle as shown below. Apply the Equilibrium of horizontal components as shown below. m v + 0 = Δ m v − Δ m u cos 20 ° + ( m − Δ m ) ( v + Δ v ) m v = Δ m v − Δ m u cos 20 ° + m v + m Δ v − Δ m v − Δ m Δ v Δ m u cos 20 ° = ( m − Δ m ) Δ v Δ v = Δ m m − Δ m u cos 20 ° Replace Δ v by d v and Δ m − d m . d v = − d m m − Δ m u cos 20 ° Here, − d m is the infinitesimal increase in m . Substitute m for ( m − Δ m ) . d v = − u cos 20 ° d m m Integrating both sides of the Equation and apply the limits as shown below

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Author: BEER

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Chapter 14.3, Problem 14.89P

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Find the top speed of the toy car.

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Chapter 14 Solutions

VECTOR MECH. FOR EGR: STATS & DYNAM (LL

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Find the top speed of the toy car.

Solution Summary: The author calculates the top speed of the toy car and the weight of water in the tank.

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