Inclined Plane Notes, Summaries of Acting

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Fg = mg= Weight angle q Parallel Force Perpendicular Force q

Inclined Plane Notes

Inclined Planes

• Weight always acts vertically
• Weight = mg= F g
• Objects move in the direction of Net Force
• Objects on an angle have a force parallel to the surface and a force perpendicular to the surface.
• Normal Force F N is always perpendicular to the surface.

Inclined Plane Questions

• Which function is used to calculate the parallel force? (sin,cos,tan) Sin
• Which function is used to calculate normal force? Cos
• What determines the size of the angle you work with? Angle of the hill
• What is the hypotenuse of the vector triangle? Weight
• What is your favorite color? Cyan

Name:_________ Label the diagram 1 - 5 are forces 6 - 8 are angles 1 2 3 4 5 6 7 8 Assuming the object is at rest Which forces are represented by Fg sinq? Which forces are represented by Fg cosq? 5 2 3 4 1 6 7 8

1. A 450 N trunk rests on a 30º inclined plane? a) What is the force acting down the plane? b) What is the force acting perpendicular to the plane? 30 o 30 o Weight Fg = mg Perpendicular Force

• A 450 N trunk rests on a 30º inclined plane?
• What is the force acting down the plane?
• Opp= hyp (sin q )= 450(sin30)= 225 N
• What is the force acting perpendicular to the plane?
• adj= hyp (cos q )= 450(cos30)= 390 N

Diagram 5 o 5 o Weight Fg = mg Perpendicular Force

Solution

• Mass = 54.7 kg
• Weight= Fg= mg= 54.7 x 9.81= 537 N
• Normal force is adjacent side
• Use Cos q = adj/hyp
• Adj = hyp (cosq) = 537 (cos 5) = 46.8 N
• Friction force is opposite side
• Use Sin q = opp/hyp
• Opp = hyp (Sinq) = 537 (sin 5) = 535 N

1. What is the force of friction holding a 225 kg box on a ramp that forms a 25º angle with the ground? 25 o 25 o Normal Force FN Weight Fg = mg Vertical Component of Weight (perpendicular force)

Solution

• Mass = 225 kg
• Weight= Fg = mg= 225 x 9.81= 2210 N
• Firction force is opposite side
• Use Sin q = opp/hyp
• Opp= hyp (Sinq) = 2210 (sin 25) = 934 N

Solution

• Mass = 520 kg
• m = F f

/ F

N

• F f is the horizontal component of the weight
• F N is the vertical component of the weight

Solution

• Weight= Fg= mg= 520 x 9.8= 5100 N
• Opp= hyp(Sinq) F f = 5100 (sin 15) = 1320 N
• F N = hyp(cosq) F N = 5100 (cos 15) = 4930 N
• F f = m F N
• m= F f

/F

N

Solution

• Mass = 125 kg
• Weight = Fg = mg = 125 x 9.81 = 1230 N
• Constant velocity means F NET

• Therefore F f

= F

parallel

• Opp = hyp (Sinq) = 1230 (sin 27) = 558 N
• F f

= 558 N