The three objects in the drawing are connected by strings that pass over massless and friction-free pulleys. The objects move, and the coefficient of kinetic friction between the middle object and the surface of the table is 0.100. (Assume

Question

m1 = 12.2 kg
and
m2 = 26.0 kg.)

Answer 1

Let me analyze the problem and write the system of equations based on the given data.

First, let's define the forces acting on each of the masses:

For m1:

1. Tension force in the string connecting it to the massless pulley, T1
2. Gravitational force, m1 * g
3. Normal force from the surface of the table, N1 (equal to m1 * g because there is no vertical acceleration)
4. Frictional force, F_friction = μ * N1 = (0.100) * (m1 * g)

For m2:

1. Tension force in the string connecting it to the massless pulley, T2
2. Gravitational force, m2 * g
3. Normal force from the surface of the table, N2 (equal to m2 * g because there is no vertical acceleration)
4. Frictional force, F_friction = μ * N2 = (0.100) * (m2 * g)

Now, let's write the equations for the accelerations of m1 and m2 based on Newton's second law of motion:

For m1 (let's consider the positive direction to be horizontally to the right):

m1 * a1 = T1 - F_friction (1)

For m2 (let's consider the positive direction to be downward):

m2 * a2 = T2 - m2 * g (2)

Since the strings and pulleys are massless and friction-free, the tension forces T1 and T2 are equal. So, T1 = T2.

Moreover, the objects have the same magnitude of acceleration as they are connected by strings. Therefore, a1 = a2 = a.

Now we can rewrite the equations (1) and (2) using these equalities:

m1 * a = T1 - (0.100) * (m1 * g) (3)

m2 * a = T1 - m2 * g (4)

We can substitute the given values of m1 and m2 into the equations (3) and (4):

12.2 * a = T1 - (0.100) * (12.2 * 9.81) (5)

26.0 * a = T1 - 26.0 * 9.81 (6)

Now we have a system of two linear equations with two unknowns, a and T1. To eliminate T1, we can multiply equation (5) by 26/12.2 and subtract equation (6):

(26.0/12.2) * (12.2 * a) = (26.0/12.2) * (T1 - (0.100) * (12.2 * 9.81)) (7)

26.0 * a = T1 - 26.0 * 9.81 (6)

Subtracting equation (6) from equation (7), we get:

- ((12.2 * 9.81) * 0.100) / (26.0/12.2) * (12.2 * a)

Solving for the acceleration, a:

a = ((12.2 * 9.81) * 0.100) / (26.0 - 12.2) = 12.012/13.8 = 0.869 m/s²

Since we found the acceleration, now we can find T1 using equation (5):

T1 = 12.2 * 0.869 + (0.100) * (12.2 * 9.81) = 20.94 N

So, the acceleration of the system is 0.869 m/s², and the tension in the string is 20.94 N.