A simple Atwood machine consists of two masses m 1 and m 2 that are connected by
ID: 1903763 • Letter: A
Question
A simple Atwood machine consists of two masses m1 and m2 that are connected by a string wound over a pulley, as seen in the figure below. Assume m2 is larger than m1. Motion in the upward direction is positive. On a piece of paper, draw two free body diagrams; one for each of the masses, showing all forces acting on each mass.
(e) Write Newton's Second Law equation (in the form Fnet = ma) for m1 in the y-direction. It is important to use the correct signs for each of the forces. Motion in the upward direction is positive and motion in the downward direction is negative. (Be sure to enter your answer in the above form with the sum of the forces in the left hand side answer box. The right hand side answer box should have the appropriate form of ma. Use the following as necessary: a, f, FN, g, m1, and T.)
T-m1g=m1a
(f) Write Newton's Second Law equation (in the form Fnet = ma) for m2 in the y-direction. It is important to use the correct signs for each of the forces. Motion in the upward direction is positive and motion in the downward direction is negative. (Be sure to enter your answer in the above form with the sum of the forces in the left hand side answer box. The right hand side answer box should have the appropriate form of ma. Use the following as necessary: a, f, FN, g, m2, and T.)
T-m2g=m2a
(g) Use your answers from parts (e) and (f) to get an expression for the magnitude of the tension T. (Use the following as necessary: m1, m2, and g.)
****I have tried T=(m1+m2)2g, AND (m1-m2)a + (m1+m2)g/2********* and got both wrong.
Explanation / Answer
The direction of weight is always down, so the direction of all mg’s is down.
Since m2 has more mass than m1, the weight of m2 is greater than the weight of m1. So 2 moves down as m1 moves up. The tension, T, is pulling m1 UP. T is up
Free body diagram for m1below
T
m1
weight = m1g
Since m1 moves up, T > m1g
Up is positive, so T is + and m2g is –
Net force = T + -m1g
T – m1g = m1 * a
acceleration will upward, so a will be positive
Free body diagram for m2 below
T
m2
weight = m2g
Since m2 moves down, m2g > T
Net force = m2g – T = m2 * a
Tension is the same, because the same rope is attached to both objects.
Solver both equations for T
T = m1g + m1 * a
T = m2g – m2 * a
set equations equal to each other
m1g + m1 * a = m2g – m2 * a
m1 * a + m2 * a = m2g – m1g
a * (m1+ m2) = m2g – m1g
a = (m2g – m1g) ÷ (m1+ m2)
a = g * (m2 – m1) ÷ (m1+ m2)
T = m1g + m1 * a
T = m1g + m1 * [g * (m2 – m1) ÷ (m1+ m2)]
T = (2 * g * m1 * m2) ÷ (m1 + m2)
-----------------
g) T = 2*g/[1/m1 + 1/m2]
h) a = g*[1 - 2/(1+m2/m1)]
i) m1: up--m2 down
j) Use the formula in (g)
k) Use the formula in (h)
l) t = [2y/a]
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