Two crates of mass m1 = 16 kg and m2 = 22 kg are connected by a cable that is strung over a pulley of mass mpulley = 24 kg as shown below. There is no friction between crate 1 and the table.
http://www.webassign.net/giocp1/8-p-058.gif
(A) Express Newton's second law for the crates (translational motion) and for the pulley (rotational motion). The linear acceleration a of the crates, the angular acceleration α of the pulley, and the tensions in the right and left portions of the rope are unknowns. (Use the following variables as necessary: m_1 for m1, m_2 for m2, alpha for α, I_pulley for Ipulley, and R_pulley for
ΣFm1=
ΣFm2=
Στ =
(B) What is the relation between a and α?
(C) Find the acceleration of the crates. (Assume that the pulley is a cylinder.)
(D) Find the tensions in the right and left portions of the rope.
Update:thats helpful but you didn't answer all the questions
Answers & Comments
Verified answer
sum of forces on m1:
T1 = m1 a where T1 is the tension in the rope connected to m1
sum of forces on m2:
T2-m2g=-m2a where T2 is the tension in the rope connected to m2
sum of torques:
(T2-T1)R = I alpha where I is the moment of inertia and alpha the angular acceleration
I for a disc = 1/2 MR^2 where M is the mass of the pulley
alpha is related to a via alpha = a/R
so the sum of torques becomes:
T2-T1=1/2Ma
using T1=m1a and T2=m2(g-a), we get:
m2(g-a)-m1a=1/2 Ma
m2g = (1/2M+m1+m2)a
a=m2g/(1/2M+m1+m2)
a=22g/(12+16+22)
a=4.31m/s/s