Applications of Newtons Laws Sample Problems Homework 1 Sample - - PDF document

Applications of Newton’s Laws

• Sample Problems
• Homework

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Sample Problem 1

In the figure below, a block B of mass M = 15.0 kg hangs by a cord from a knot K of mass mK, which hangs from the ceiling by means of two other cords. The cords have negligible mass, and the magnitude of the gravitational force on the knot is negligible compared to the gravita- tional force on the block. What are the tensions in the three cords?

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Sample Problem 1 (cont’d)

Block ⇒

Fy = T3 − Fg = 0

T3 = Fg = Mg = (15.0 kg)

• 9.8 m/s2
• = 147 N

Knot ⇒

Fx = T2 cos 47◦ − T1 cos 28◦ = 0

T2 = cos 28◦ cos 47◦T1 = 1.29T1

Fy = T1 sin 28◦ + T2 sin 47◦ − T3 = 0

T1 sin 28◦ + (1.29T1) sin 47◦ − T3 = 0 T1 = T3 sin 28◦ + 1.29 sin 47◦ T1 = 147 N sin 28◦ + 1.29 sin 47◦ = 104 N T2 = 1.29T1 = 1.29 (104 N) = 134 N

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Sample Problem 2

In the figure below, a cord holds stationary a block of mass m = 15 kg, on a frictionless plane that is inclined at an angle θ = 27◦. What are the magnitudes of the tension in the cord and the normal force on the block?

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Sample Problem 2 (cont’d)

Fx = T − Fg sin θ = 0

T = Fg sin θ = mg sin θ T = (15 kg)

• 9.8 m/s2
• sin 27◦ = 67 N

Fy = N − Fg cos θ = 0

N = Fg cos θ = mg cos θ N = (15 kg)

• 9.8 m/s2
• cos 27◦ = 130 N

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Sample Problem 3

In the figure below, a passenger of mass m = 72.2 kg stands on a platform scale in an elevator. (a) Find a gen- eral expression for the scale reading, whatever the ver- tical motion of the cab. (b) What does the scale read if the elevator is stationary or moving upward at a constant speed of 0.50 m/s? (c) What does the scale read if the elevator accelerates upward at 3.20 m/s2 and downward at 3.20 m/s2?

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Sample Problem 3 (cont’d)

(a) Find a general expression for the scale reading, what- ever the vertical motion of the cab.

Fy = N − Fg = may

N = Fg + may = mg + may = m (g + ay) (b) What does the scale read if the elevator is stationary

• r moving upward at a constant speed of 0.50 m/s?

N = (72.2 kg)

• 9.8 m/s2 + 0
• = 708 N

(c) What does the scale read if the elevator accelerates upward at 3.20 m/s2 and downward at 3.20 m/s2? Up ⇒ N = (72.2 kg)

• 9.8 m/s2 + 3.20 m/s2
• = 939 N

Down ⇒ N = (72.2 kg)

• 9.8 m/s2 − 3.20 m/s2
• = 477 N

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Sample Problem 4

Consider Atwood’s machine shown in the figure. As- sume m2 > m1 and the masses of the cord and the pulley are negligible. Find an expression for the acceleration.

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Sample Problem 4 (cont’d)

Fm1 = T − m1g = m1a Fm2 = m2g − T = m2a

Adding these two eqns. we get m2g − m1g = m1a + m2a a = m2 − m1 m1 + m2 g

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Homework Set 8 - Due Mon. Sept. 27

• Read Section 4.7
• Do Problems 4.22, 4.25, 4.26, 4.30, 4.31 & 4.34

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