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Aim: How can we solve problems dealing with Atwood Machines using Newton’s Laws? HW #6 Do Now: An object is of mass M is hanging from a rope as shown.

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Presentation on theme: "Aim: How can we solve problems dealing with Atwood Machines using Newton’s Laws? HW #6 Do Now: An object is of mass M is hanging from a rope as shown."— Presentation transcript:

1 Aim: How can we solve problems dealing with Atwood Machines using Newton’s Laws?
HW #6 Do Now: An object is of mass M is hanging from a rope as shown. Calculate its acceleration as it drops. T ΣF = ma mg – T = ma M M mg

2 Atwood Machines Invented in 1784 by Rev. George Atwood
Used to verify the mechanical laws of uniformly accelerated motion. The ideal Atwood Machine consists of two objects of mass m1 and m2, connected by a string over a pulley. When m1 = m2, the machine is in neutral equilibrium regardless of the position of the weights. When m2 > m1 both masses experience uniform acceleration M2 m1

3 Each mass gets its own free-body diagram
Positive is denoted by the direction of acceleration Add the two equations together to cancel out tension Solve for acceleration Substitute acceleration into one of the formulas to solve for tension

4 1. A block of mass 3m can move without friction on a horizontal table
1. A block of mass 3m can move without friction on a horizontal table. This block is attached to another block of mass m by a cord that passes over a frictionless pulley, as shown above. If the masses of the cord and the pulley are negligible, what is the magnitude of the acceleration of the descending block? (A) Zero (B) g/4 (C) g/3 (D) 2g/3 (E) g 3m T T = 3ma mg - T = ma mg = 3ma + ma mg = 4ma g = 4a a = g/4 ΣF = ma T = 3ma T m ΣF = ma mg - T = ma No Calculator **1 min 15 sec** mg

5 Calculator Allowed **15 min** 2. Two small blocks, each of mass m, are connected by a string of constant length 4h and negligible mass. Block A is placed on a smooth tabletop as shown above, and block B hangs over the edge of the table. The tabletop is a distance 2h above the floor. Block B is then released from rest at a distance h above the floor at time t = 0. Express all algebraic answers in terms of h, m, and g.

6 a. Determine the acceleration of block A as it descends.
mg mg ΣF = ma T = ma ΣF = ma mg - T = ma T = ma mg – T = ma mg = ma + ma g = a + a g = 2a a = g/2

7 b. Block B strikes the floor and does not bounce
b. Block B strikes the floor and does not bounce. Determine the time t = t1 at which block B strikes the floor.

8 Describe the motion of block A from time t = 0 to the time when block B strikes the floor.
Block A accelerates across the table Describe the motion of block A from the time block B strikes the floor to the time block A leaves the table. Block A remains in motion with constant velocity

9 e. Determine the distance between the landing points of the two blocks.
Block B falls straight to the floor. It’s velocity when it hits is equal to the horizontal velocity of block A as it leaves the table. Block A travels as a horizontally fired object as it leaves the table

10 3. Three blocks of masses 1. 0, 2. 0, and 4
3. Three blocks of masses 1.0, 2.0, and 4.0 kilograms are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above. Calculate each of the following. a. The acceleration of the 4‑kilogram block b. The tension in the string supporting the 4‑kilogram block c. The tension in the string connected to the l‑kilogram block Calculator Allowed **10 min**

11 ΣF = ma T – mg = ma T – (3 kg)(9.8 m/s2) = (3 kg)a T – 29.4 = 3a
mg - T = ma (4 kg)(9.8 m/s2) – T = (4 kg)a 39.2 –T = 4a T – 29.4 = 3a 39.2 –T = 4a 9.8 = 7a a = 1.4 m/s2

12 T – 29.4 = 3a T – 29.4 = 3(1.4 m/s2) T – 29.4 = 4.2 T = 33.6 N ΣF = ma
c. T b. T – 29.4 = 3a T – 29.4 = 3(1.4 m/s2) T – 29.4 = 4.2 T = 33.6 N 1 kg a mg ΣF = ma T – mg = ma T – (1 kg)(9.8 m/s2) = (1 kg)(1.4 m/s2) T – 9.8 = 1.4 T = 11.2 N


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