An Atwood\'s machine consists of blocks of masses m 1 = 11.0 kg and m 2 = 22.0 k

Question

An Atwood's machine consists of blocks of masses m1 = 11.0 kg and m2 = 22.0 kg attached by a cord running over a pulley as in the figure below. The pulley is a solid cylinder with mass M = 7.00 kg and radius r = 0.200 m. The block of mass m2 is allowed to drop, and the cord turns the pulley without slipping.

This answer has not been graded yet.

T1 = N T2 = N An Atwood's machine consists of blocks of masses m1 = 11.0 kg and m2 = 22.0 kg attached by a cord running over a pulley as in the figure below. The pulley is a solid cylinder with mass M = 7.00 kg and radius r = 0.200 m. The block of mass m2 is allowed to drop, and the cord turns the pulley without slipping. Why must the tension T2 be greater than the tension T1? What is the acceleration of the system, assuming the pulley axis is frictionless? Find the tensions T1 and T2.

Explanation / Answer

For Tension 2:
g=9.8m/s^2

m2(9.8)-T2=m2(a)
** 22(9.8)-T2=22(a)

T2=

For Tension 1:

T1-m1(9.8)=m1(a)
** T1-11(9.8)=11(a)

For solid cylinder:
r=radius
I=moment of inertia
alpha=angular acceleration
a=acceleration

rT2-T1= I*alpha
I=.5mr^2
alpha=a/r
r(T2-T1)= .5mr^2(a/r)

radius cancels out leaving:

T2-T1=.5(mass of cylinder)a=
** T2-T1=.5(7.70)= 3.85(a)

Tensions cancel out, Now you add up all the equations with an asterisk next to them:

this leaves 21.1g-10.8g=35.75(a)
g is just gravity so multiply 9.8 by the masses then (a) will be your only unknown

after finding (a), plug it back into the 1st equation with an asterisk and solve for T2, then plug into second equation, solve for T1 and you're done

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.