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1) If drag is 26,400 N and the lift is 29,300 N, determine the weight, mass and thrust, given that the plane is traveling at a constant velocity. What.

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Presentation on theme: "1) If drag is 26,400 N and the lift is 29,300 N, determine the weight, mass and thrust, given that the plane is traveling at a constant velocity. What."— Presentation transcript:

1 1) If drag is 26,400 N and the lift is 29,300 N, determine the weight, mass and thrust, given that the plane is traveling at a constant velocity. What is the net force acting on the plane? 2) If the net force acting on the plane is 9250 N to the left, and the thrust force is 35,000 N, what is the drag force? What is the rate of acceleration? Fnet = ma Weight = Fgrav = mg Ch. 4 HW Dec 11

2 Draw Free Body Diagrams 1.0.5 N acorn falling to the ground with air resistance totaling ¼ of its weight. – What is the net force acting on the acorn? Is the acorn accelerating? If so, at which rate? 2.A 45 N box being pushed to the right with 35 N of force, with a net force of 9 N to the right. – Is the box accelerating? If so, at which rate and in which direction? 3.A 1.5 kg book resting horizontally on a table 4.Draw the four forces acting on the book if it is at rest on a 15 degree slope. Fg,y = Fn = Fg cos Fg,x = Fg sin represents the angle of the slope

3 26,400 N 29,300 N 26,400 N m = W/g = 29,300 N/9.81 m/s 2 = 2990 N Net Force = 0 N Drag force = 25,750 N 1) If drag is 26,400 N and the lift is 29,300 N, determine the weight, mass and thrust, given that the plane is traveling at a constant velocity. What is the net force acting on the plane? 2) If the net force acting on the plane is 9250 N to the left, and the thrust force is 35,000 N, what is the drag force? What is the rate of acceleration? a = Fnet/m = 9250 N/2990 kg = 3.09 m/s 2

4 F air = 0.125 N Fg = 0.5 N m=w/g= 0.5/9.81 m/s 2 = 0.510 kg F applied = 35 N Fg = 40 N F norm = 40 N F friction = 26 N F net = 9 N to the right Fg = 15 N (w = 1.5 kg x 9.8 m/s 2 ) F norm = 15 N Free Body Diagram Practice F net =.375 N downward a = Fnet/m = 0.375N/.0510 kg = 7.35 m/s 2 downward m = 40 N /9.81 m/s 2 = 4 kg a = Fnet/m = 9 N/5 kg = 2 m/s 2 to the right

5 Rounded to two sig figs Fg,x = 3.9 N Fg = 15 N Fn = 15 N Fs = 3.9 N Fg,y = 14 N 15°


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