Yusef pushes a chair of mass m= 50.0 kgacross a carpeted floor with a force of m

Question

Yusef pushes a chair of mass  m= 50.0 kgacross a carpeted floor with a force of magnitudeFP = 160 N directed at = 35.0 degrees below thehoriztonal . The magnitude of the frictional force between thecarpet and the floor isFFR = 98.9 N. 1. Use the component form of Newton's second lawto write an expression for the x component of the netforce, Fx Express your answer in terms of someor all of the variables: FG, FN,FP, , and FFr . 2. Use the component form of Newton's second lawto write an expression for the y component of the netforce, Fy Express your answer in terms of someor all of the variables: FG, FN,FP, , and FFr . Yusef pushes a chair of mass  m= 50.0 kgacross a carpeted floor with a force of magnitudeFP = 160 N directed at = 35.0 degrees below thehoriztonal . The magnitude of the frictional force between thecarpet and the floor isFFR = 98.9 N. 1. Use the component form of Newton's second lawto write an expression for the x component of the netforce, Fx Express your answer in terms of someor all of the variables: FG, FN,FP, , and FFr . 2. Use the component form of Newton's second lawto write an expression for the y component of the netforce, Fy Express your answer in terms of someor all of the variables: FG, FN,FP, , and FFr . Use the component form of Newton's second lawto write an expression for the x component of the netforce, Fx Express your answer in terms of someor all of the variables: FG, FN,FP, , and FFr . 2. Use the component form of Newton's second lawto write an expression for the y component of the netforce, Fy Express your answer in terms of someor all of the variables: FG, FN,FP, , and FFr . Use the component form of Newton's second lawto write an expression for the y component of the netforce, Fy Express your answer in terms of someor all of the variables: FG, FN,FP, , and FFr .

Explanation / Answer

FP = 160 N X-component F p = 160 cos                             = 131 N
Angle = 35.0 degrees mass m = 50 kg magnitude of the frictional force between the carpet and thefloor is FFR = 98.9 N (1) . the x component of the net force,Fx = 131 - 98.9                                                                 = 32.1 N (2).  y component of the net force,Fy = 0 Since normal force N = mg - F sin

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