Two masses (m1 and m2) are hanging from a cord, threaded through a massless pull

Question

Two masses (m1 and m2) are hanging from a cord, threaded through a massless pulley (similar to figure 4-46, p 108). The pulley is hanging from the ceiling by cord C. In addition, an additional third mass (m3) is hung from mass 1 by an additional cord. a. Derive an expression for the acceleration of mass 1 in terms of m1, m2, m3, and g. Make sure to start with the FBD for each of the three masses plus the pulley, and then start from first principles. b. If m1=1.0kg, m2=3.0kg, and m3=7.0kg, calculate the tension in cord C. c. Re-derive the acceleration, and the tension in cord C if mass 3 is instead hung from mass 2. You may start from first principles, or you may use symmetry arguments to compare with parts a and b.

If anyone can help, I would greatly appreciate it. I came up with an answer, but I want to see what some of whiz kids can come up with! Thanks!

Explanation / Answer

here,
By using Newton second law of motion,

Fnet for Mass M1
m1g +m3g - Fnet = M1*A1
(g(M1 + M3) - T ) / (M1+M3) = A-----------------------(1)

Fnet for mass M2

Fnet - M*2g = M2*A

T - M*2g = M2*A
(T - M2g)/M2 = A ----------------------(2)

as M1 = 1 kg
M2 = 3 kg
M3 = 7 kg

using these value in 1 and 2 we get

( 9.8(1 + 7) - T ) / 7 + 1 = A

and

(T - 3*9.8 )/3 = A

rearranging and using equating for solving T

( 78.4 -T) /8 = (T - 29.4)/3

T = 42.76 N

therefore

Fnet = (m1+m3)a

42.76 = 8a

a = 5.34 m/s^2

Part C:

m1g - Fnet = M1*A1
(g(M1) - T ) / (M1) = A-----------------------(1)

Fnet for mass M2

Fnet - (M2+M3)g = (M2+M3)A

T - (M2+M3)g = M2*A
(T - (M2+M3)g)/(M2+M3) = A ----------------------(2)

as M1 = 1 kg
M2 = 3 kg
M3 = 7 kg

using these value in 1 and 2 we get

(9.8*1 - T ) / 1 = A
(T - 98) / 10 =A
rearranging and using equating for solving T

9.8 - T = (T - 98 ) /10

T = 17.81 N

therefore

17.81 = (M2+M3) = 10a

a = 17.81 / 10 = 1.781 m/s^2

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