Two crates of mass m1 = 16 kg and m2 = 23 kg are connected by a cable that is strung over a pulley of mass mpulley = 22 kg as shown below. There is no friction between crate 1 and the table.

m1 is sitting on the table
mpulley is at the edge of the table to the right of m1
m2 is hanging off the table going straight down
there is a rope between m1 and m2 as well as the pulley-the rope going through the pulley then to m2

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 á 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 á, I_pulley for Ipulley, and R_pulley for Rpulley.)
ÓFm1 =
ÓFm2 =
Óô =

What is the relation between a and á?
a =

Find the acceleration of the crates. (Assume that the pulley is a cylinder.)
m/s2

Find the tensions in the right and left portions of the rope.
Left Side N
Right Side N

To solve this problem, we can start by writing Newton's second law for each object involved - crate 1, crate 2, and the pulley.

For crate 1 (m1) on the table, there is no friction, so the only force acting on it is its weight (mg1). Therefore, Newton's second law for crate 1 is given by:

ΣFm1 = m1 * a = m1 * g (1)

Here, a represents the linear acceleration of crate 1 and g represents the acceleration due to gravity.

For crate 2 (m2) hanging off the table, the tension in the rope (T) and its weight (mg2) are the forces acting on it. Newton's second law for crate 2 is:

ΣFm2 = m2 * g - T = m2 * a (2)

Here, a represents the linear acceleration of crate 2.

Now, let's move on to the pulley. Since it is a cylinder, it experiences both translational motion (linear acceleration) and rotational motion (angular acceleration). The net force (ΣFpulley) acting on the pulley is the difference between the tensions on the left and right sides of the rope (TL - TR).

ΣFpulley = TL - TR (3)

The net torque (Στ) acting on the pulley can be calculated from the moment of inertia of the pulley (Ipulley) and its angular acceleration (α).

Στ = Ipulley * α (4)

Since the rope is inextensible and does not slip over the pulley, we know that the linear acceleration of crate 1 (a) and the angular acceleration of the pulley (α) are related by the equation:

a * Rpulley = α * Rpulley (5)

Where, Rpulley represents the radius of the pulley.

Now, let's solve the equations to find the unknowns in this problem:

To find the relation between a and α, we can use equation (5):

a = α (6)

Therefore, the relation between a and α is that they are equal.

To find the acceleration of the crates, we need to solve equations (1) and (2) simultaneously. Adding equation (1) and equation (2), we get:

m1 * a + m2 * a = m1 * g + m2 * g

(a * (m1 + m2)) = g * (m1 + m2)

a = g

Since the mass terms cancel out, the acceleration of the crates (a) is equal to the acceleration due to gravity (g). Therefore, the acceleration of the crates is g = 9.8 m/s^2.

To find the tensions in the right and left portions of the rope (TR and TL), we can substitute the value of a (g = 9.8 m/s^2) into equation (2):

m2 * g - T = m2 * a

m2 * g - T = m2 * 9.8

Solving for T, we get:

T = m2 * g - m2 * 9.8

On the left side of the rope, the tension (TL) is equal to the weight of crate 1:

TL = m1 * g

Therefore, the tensions in the right and left portions of the rope are:

Left Side TL: m1 * g
Right Side TR: m2 * g - m2 * 9.8

Now, you can substitute the given values for masses (m1 = 16 kg, m2 = 23 kg) and acceleration due to gravity (g = 9.8 m/s^2) to calculate the numerical values.