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Chapter_5_09

# Chapter_5_09 - Chapter 5 Problems 5.3 Only two horizontal...

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Chapter 5 Problems 5.3 Only two horizontal forces act on a 3.0 kg body. One force is 9.0 N, acting due east , and the other is 8.0N act 62 degrees North of west. What is the magnitude of the body’s acceleration? F 1 = 9 ˆ i + 0 ˆ j F 2 = 8cos62 ° ˆ i + 8sin62 ° ˆ j F net = (9 ° ) ˆ i + (8sin62 ° ) ˆ j = 5.244 ˆ i + 7.066 ˆ j F net = 5.244 2 + 7.066 2 = 8.8 N a = F net m = 8.8 N 3 kg = 2.93 m / s 2 5.7. Three astronauts, propelled by jet backpacks, push and a guide a 120 kg asteroid toward a processing dock, exerting the forces shown in Fig. 5-30. What is the asteroid’s acceleration (a) in unit vector notation and as (b) a magnitude and (c) a direction. 30˚ 60˚ 32N 55N 41N a) We begin this problem by writing out each force in component form and then writing the net force.

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F 1 = 32 cos30 ° ˆ i + 32sin 30 ° ˆ j F 2 = 55 ˆ i + 0 ˆ j F 3 = 41cos60 ° ˆ i 41sin 60 ° ˆ j F net = (32 cos30 ° + 55 + 41cos60 ° ) ˆ i + (32sin 30 ° + 0 41sin60 ° ) ˆ j = 118.22 ˆ i 19.51 ˆ j Now that we know the net force, we can Fnd the acceleration. a = F net m = 118.22 ˆ i 19.51 ˆ j N 120 kg = 0.985 ˆ i 0.163 ˆ j The magnitude and direction are: a = 0.985 2 + 0.163 2 = 0.998 m / s 2 tan θ = 0.163 0.985 = 9.4 ° 5.14 A block with a weight of 3.0 N is at rest on a horizontal surface. A 1.0 N upward force is applied to the block by means of an attached vertical string. What are the (a) magnitude and (b) direction of the force of the block on the horizontal surface. The force of the block on the surface is equal and opposite to the force of surface on the block (the normal force). This problem is thus really about Fnding the normal force. W T N N T W We can now write the net force in the y direction and use it to compute N. 0
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Chapter_5_09 - Chapter 5 Problems 5.3 Only two horizontal...

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