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A boy pulls horizontally with a force of 100N on a box of Mass 50kg. Find the resulting acceleration if (a) Force of friction = 0; (b) Force of friction.

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Presentation on theme: "A boy pulls horizontally with a force of 100N on a box of Mass 50kg. Find the resulting acceleration if (a) Force of friction = 0; (b) Force of friction."— Presentation transcript:

1 A boy pulls horizontally with a force of 100N on a box of Mass 50kg. Find the resulting acceleration if (a) Force of friction = 0; (b) Force of friction = 25N. F appl FgFg FNFN F fr + + - -

2 A boy pulls horizontally with a force of 100N on a box of Mass 50kg. Find the acceleration motion if (a) Force of friction = 0; (b) Force of friction = 25N. F appl FgFg FNFN F fr + + - - (a) Fx = Fappl – Ffr = ma 100N = (50kg)a a = 2 m/s 2 Fy = F N – Fg = ma = 0 (there is no motion or acceleration in the y-direction)

3 A boy pulls horizontally with a force of 100N on a box of Mass 50kg. Find the acceleration motion if (a) Force of friction = 0; (b) Force of friction = 25N. F appl FgFg FNFN F fr + + - - (b) Fx = Fappl – Ffr = ma 100N – 25N = (50kg)a a = 1.5 m/s 2 NOTE: Fy = F N – Fg = ma = 0 F N = (50kg)(9.8)=490N (there is no motion or acceleration in the y-direction)

4 A boy pulls at a 30˚ angle with a force of 100N on a box of Mass 50kg. Find the resulting acceleration if force of friction = 0. F appl FgFg FNFN F fr + + - -

5 A boy pulls at a 30˚ angle with a force of 100N on a box of Mass 50kg. Find the resulting acceleration if force of friction = 0. F appl FgFg FNFN F fr + + - - Fx = Fappl cos30 – Ffr = ma 100N cos30 = (50kg)a a = 1.73 m/s 2 NOTE: Fy = F N + Fappl sin30 – Fg = ma F N + 100 sin30 – (50)(9.8) = 0 (there is no motion or acceleration in the y-direction) F N = -50 + 490 = 440N

6 A boat is pulled with two forces as shown (one at a 45˚ angle and one at a 37 ˚ angle. Calculate the net force on the boat. + + 45˚ 37˚ F1 = 40N F2 = 30N

7 A boat is pulled with two forces as shown (one at a 45˚ angle and one at a 37 ˚ angle. Calculate the net force on the boat. + + 45˚ 37˚ F1 = 40N F2 = 30N F1x = F1cos45 = 28.3 N F1y = F1sin45 = 28.3N F2x = F2cos37 = 24.0N F2y = F2sin37 = -18.1N Fx = F1x + F2x = 52.3N Fy = F1y + F2y = 10.2N F = (Fx 2 + Fy 2 ) 1/2 = 53.3N tanΘ = 10.2/52.3 Angle is 11.0˚

8 When a flexible cord pulls on an object (assume the cord has Negligible mass), calculate the acceleration of the system. The Pulling force is 40N and the boxes have a mass of 10kg each. F appl FgFg FNFN + + - FgFg FNFN + - FTFT FTFT

9 When a flexible cord pulls on an object (assume the cord has Negligible mass), calculate the acceleration of the system. The Pulling force is 40N and the boxes have a mass of 10kg each. m1 F appl FgFg FNFN + + - m2 FgFg FNFN + - FTFT FTFT Fx=F T =m 2 a Fx=Fappl - F T =m 1 a Fappl – m 2 a = m 1 a NOTE: a is the same for both boxes. a= 2 m/s 2

10 You are given an Atwood’s Machine. There are two masses hanging from a massless pulley. M1 is 15kg and M2 is 28kg. Calculate acceleration of the masses and the force of tension in the rope. m1 F g1 + m2 F g2 - FTFT FTFT

11 You are given an Atwood’s Machine. There are two masses hanging from a massless pulley. M1 is 15kg and M2 is 28kg. Calculate acceleration of the masses and the force of tension in the rope. m1 F g1 + m2 F g2 - FTFT FTFT F T – Fg 2 = m 2 a -F T + Fg 1 = m 1 a NOTE: a is the same for both masses. a= -2.96 m/s 2 a is to the left mass. + - Fg 1 – Fg 2 = (m1 + m2)a (15kg)(9.8) – (28)(9.8) = (43kg)(a)


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