Need help, mechanics problem with pulleys and tensions

  • #1
I've already tried this problem but yielded an impossible answer:

Homework Statement



We have a pulley with a mass-less rope attached to two masses hanging off each side of the pulley:
-m1 (4kg), suspended 1m off the ground
-m2 (2kg), on the ground

The mass (M) of the pulley is 2.6 kg
The rotational friction of the pulley is given as .51 Nm
The radius of the pulley is .085

How long will it take for the 4kg mass to hit the ground once the masses are released?

Homework Equations



Moment of inertia (I) = (1/2)(mass)(radius)2
Torque (τ) = (moment of inertia)(angular acceleration)
Angular acceleration (α) = acceleration of point on outside edge of pulley / radius

The Attempt at a Solution



I defined moving the pulley so that m1 falls as "positive" rotation and angular acceleration.

Regarding mass one,
Fnet = m1g + T1 = ma
T1 = m1a - m1g

For mass two,
Fnet = - m2g + T2 = ma
T2 = m2a + m2g

τnet = I α
τ(from tension) - τ(from friction) = ((1/2) M r2) (a / r)
T1r - T2r - .51 = (1/2)(2.6)(r)(a)
T1 - T2 - .51/r = (1/2)(2.6)(a)
m1a - m1g - m2a - m2g = 1.3a
a(m1 - m2 - 1.3) = m1g + m2g
a(4 - 2 - 1.3) = 39.2 + 19.6
a = 84

An acceleration greater than the acceleration due to gravity would be impossible for mass one.

Would any please check what I did wrong? Thank you!
 

Answers and Replies

  • #2
Doc Al
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Regarding mass one,
Fnet = m1g + T1 = ma
T1 = m1a - m1g
Check the sign of the tension.
 
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  • #3
-m1a + m1g - m2a - m2g = 1.3a
a(-m1 - m2 - 1.3) = -m1g + m2g
a(-4 - 2 - 1.3) = -39.2 + 19.6
a = 2.68

That was the problem. Thanks!
 

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