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Finding acceleration of a three-object pulley system

  • Thread starter kathyt.25
  • Start date
  • #1
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Homework Statement


"Three blocks are connected on a table as shown in the figure below.
http://i4.photobucket.com/albums/y111/kathy_felldown/sb-pic0550.png

The table is rough and has a coefficient of kinetic friction of 0.360. The three masses are m1 = 3.74 kg, m2 =1.39 kg, and m3 = 1.86 kg, and the pulleys are frictionless. Determine the magnitude of the acceleration of each block."

I think I've figured out how to do it, but I keep on getting the wrong answer. I denoted left to be the POSITIVE x-direction, and down to be the POSITIVE y-direction, since that's the direction of movement. Can someone check my equations to see if I have them right?


Homework Equations


Fnet=ma
f(k)=u*n (where u=coefficient of friction)


The Attempt at a Solution


For m1, the 3.74kg mass:
Fnet(y) = m1*g - T1 = ma
Therefore, T1 = m1*g - m1*a --> EQUATION #1

For m2, the 1.39kg mas on the table:
Fnet(y) = m2*g - n = 0
Therefore, n = m2*g

Fnet(x) = T1 - T3 - f = ma
Therefore, T1 - T3 - u*n = ma --> EQUATION #2

For m3, the 1.86kg mass:
Fnet(y) = -T3 +m3*g = ma
Therefore, T3 = m3*g - m3*a --> EQUATION #3


I then substituted equations #1 and #3, into equations #2, and isolated for acceleration "a".
 

Answers and Replies

  • #2
Doc Al
Mentor
44,882
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For m3, the 1.86kg mass:
Fnet(y) = -T3 +m3*g = ma
Therefore, T3 = m3*g - m3*a --> EQUATION #3
You've made a sign error. The acceleration of m3 is upward and must have the appropriate sign.
 
  • #3
181
0
If m1 has acceleration a downwards, the acceleration of m3
is a upwards (or -a downwards if you prefer)
 

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