Choosing right to be the positive direction we 1 2 2 cos 590 N

Choosing right to be the positive direction we 1 2 2

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Choosing right to be the positive direction, we have(29(29122cos59.0 Ncos70.033.0 N1.83 m/s7.00 kgxxxFFFammaθΣ-==° -== -The minus sign indicates that the horizontal acceleration points to the left.12.REASONINGThe net force ΣFhas a horizontal component ΣFxand a vertical component ΣFy. Since these components are perpendicular, the Pythagorean theorem applies (Equation 1.7), and the magnitude of the net force is (29(2922xyFFFΣ=Σ+ Σ. Newton’s second law allows us to express the components of the net force acting on the ball in terms of its mass and the horizontal and vertical components of its acceleration: , xxyyFmaFmaΣ=Σ=(Equations 4.2a and 4.2b). SOLUTIONCombining the Pythagorean theorem with Newton’s second law, we obtain the magnitude of the net force acting on the ball: (29(29(29(29(29(29(2922222222220.430 kg810 m/s1100 m/s590 NxyxyxyFFFmamamaaΣ=Σ+ Σ=+=+=+=+y+xFNF2F1Wθ+y+xFNF2F1cos θWF1sin θ
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168FORCES AND NEWTON'S LAWS OF MOTION13.SSMREASONINGTo determine the acceleration we will use Newton’s second law ΣF= ma. Two forces act on the rocket, the thrust Tand the rocket’s weight W, which is mg = (4.50 × 105kg)(9.80 m/s2) = 4.41 × 106N. Both of these forces must be considered when determining the net force ΣF. The direction of the acceleration is the same as the direction of the net force.SOLUTIONIn constructing the free-body diagram for the rocket we choose upward and to the right as the positive directions. The free-body diagram is as follows:The xcomponent of the net force is(2966cos55.07.5010Ncos55.04.3010NxFTΣ=°=×° =×The ycomponent of the net force is(29666sin55.07.5010Nsin55.04.4110N1.7310NyFTWΣ=°-=×° -×=×The magnitudes of the net force and of the acceleration are(29(29(29(29(29(2922222266254.3010N1.7310N10.3 m/s4.5010kgxyxyFFFFFamΣ=Σ+ ΣΣ+ Σ×+×===×The direction of the acceleration is the same as the direction of the net force. Thus, it is directed above the horizontal at an angle of61161.7310Ntantan21.94.3010NyxFFθ--Σ×===°÷÷÷÷Σ×14.REASONINGThe acceleration of the sky diver can be obtained directly from Newton’s second law as the net force divided by the sky diver’s mass. The net force is the vector sum of the sky diver’s weight and the drag force.SOLUTION From Newton’s second law, FmaΣ=(Equation 4.1), the sky diver’s acceleration isFamΣ=55.0º+x+yTWTx= Tcos 55.0ºTy= Tsin 55.0ºfWFree-body diagram+-
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Chapter 4 Problems 169The free-body diagram shows the two forces acting on the sky diver, his weight Wand the drag force f. The net force is FfWΣ=-. Thus, the acceleration can be written asfWam-=The acceleration of the sky diver is21027 N915 N1.20 m/s93.4 kgfWam--=== +Note that the acceleration is positive, indicating that it points upward .
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