Two crates of mass m 1 = 12 kg and m 2 = 22 kg are connected by a cable that is

Question

Two crates of mass m1 = 12 kg and m2 = 22 kg are connected by a cable that is strung over a pulley of mass mpulley = 17 kg as shown in the figure below. There is no friction between crate 1 and the table.

(b) 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 ? of the pulley, and the tensions in the vertical and horizontal portions of the rope are unknowns. (Assume the positive direction is to the right of m1 and downward from m2. Assume Rpulley is the radius of the pulley, Ipulley is the moment of inertia of the pulley, g is gravitational acceleration, and T1 and T2 are the horizontal and vertical portions of the tension, respectively.)

For the left crate:

A. ?Fm1 = T1 = m1a

B. ?Fm1 = ?T2 + m2g = m1a   

C .?Fm1 = T2Rpulley ? T1Rpulley = m1a

D. ?Fm1 = ?T1 + m2g = m1a

For the right crate:

A. ?Fm2 = T2Rpulley ? T1Rpulley = m2a

B. ?Fm2 = ?T2 + m2g = m2a    

C. ?Fm2 = ?T1 + m2g = m2a

D. ?Fm2 = ?m1g + T1 = m2a

For the pulley:

A. ?? = ?m1g + T1 = Ipulley?

B. ?? = ?T1 + m2g = Ipulley?    

C. ?? = T2Rpulley ? T1Rpulley = Ipulley?

D. ?? = ?T2 + m2g = Ipulley?

(c) What is the relation between a and ??

a = 2??Rpulley

a =

    a =

a = ?Rpulley

(d) Find the acceleration (magnitude only) of the crates.
m/s2

(e) Find the tensions in the horizontal and vertical portions of the rope.

Please do the best you can. I don't mind if you show some parts. Thanks!

Explanation / 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/(17/2 +12+22)
a= 5.078 m/s^2

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