Two packing crates of masses m1 = 10.0 kg and m2 = 6.50 kg are connected by a light string that passes over a frictionless pulley as in Figure P4.26. The 6.50 kg crate lies on a smooth incline of angle 39.0°.

a)Find the acceleration of the 6.50 kg crate.
m/s2 (up the incline)

b)Find the tension in the string.
N

m1 and m2 will have the same acceleration rate magnitude, but it will be in difference directions.

Draw free body diagrams of m1 and m2 write two simultaneous equations of motion (one for each mass). The unknowns will be string tension T and acceleration rate a.

Someone will be glad to critique your work.

13.8

To solve this problem, we can break it down into two parts: the motion of the 10.0 kg crate and the motion of the 6.50 kg crate.

First, let's analyze the motion of the 10.0 kg crate. Since there is no friction and the string is light, the tension in the string will be the same on both sides. Therefore, the weight of the 10.0 kg crate will be balanced by the tension in the string:

T - m1 * g = 0

where T is the tension in the string, m1 is the mass of the 10.0 kg crate, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Now, let's analyze the motion of the 6.50 kg crate. The weight of this crate can be separated into two components: one parallel to the incline and one perpendicular to the incline. The component parallel to the incline will be responsible for the acceleration of the crate, while the component perpendicular to the incline will be balanced by the normal force.

The component parallel to the incline is given by:
m2 * g * sin(theta)

where m2 is the mass of the 6.50 kg crate and theta is the angle of the incline (39.0°).

Since the tension in the string is the same on both sides, the acceleration of the 6.50 kg crate is equal to the component parallel to the incline divided by the mass of the 6.50 kg crate:

a = (m2 * g * sin(theta)) / m2

Now we can calculate the values:

a) We plug in the given values:
a = (6.50 kg * 9.8 m/s^2 * sin(39.0°)) / 6.50 kg

a ≈ 4.04 m/s^2 (up the incline)

b) To find the tension in the string, we use the equation:
T = m1 * g - m2 * g * sin(θ)

T = (10.0 kg * 9.8 m/s^2) - (6.50 kg * 9.8 m/s^2 * sin(39.0°))

T ≈ 34.4 N

So, the acceleration of the 6.50 kg crate up the incline is approximately 4.04 m/s^2, and the tension in the string is approximately 34.4 N.

To find the acceleration of the 6.50 kg crate on the incline, we need to analyze the forces acting on it.

First, let's draw a free body diagram for the 6.50 kg crate. The force of gravity acting vertically downward can be split into two components: one parallel to the incline and another perpendicular to the incline. The perpendicular component doesn't affect the motion along the incline, so we can ignore it for now.

The parallel component of the force of gravity is given by:

F_parallel = m2 * g * sin(θ)
where m2 is the mass of the 6.50 kg crate, g is the acceleration due to gravity (approximately 9.8 m/s^2), and θ is the angle of the incline (39.0°).

Now, we can use Newton's second law of motion to find the acceleration, a.

Sum of the forces along the incline = m2 * a

The only force along the incline is the component of the force of gravity, so we have:

F_parallel = m2 * a

Plugging in the values we know:

m2 * g * sin(θ) = m2 * a

Now, we can divide both sides of the equation by m2:

g * sin(θ) = a

Plugging in the values:

a = 9.8 m/s^2 * sin(39.0°)

Calculating the value:

a ≈ 5.98 m/s^2 (up the incline)

Therefore, the acceleration of the 6.50 kg crate on the incline is approximately 5.98 m/s^2 (up the incline).

Now, let's move on to finding the tension in the string. The tension in the string is the same on both sides of the pulley.

For the 10.0 kg crate, the net force acting on it is the force of gravity pulling it downward:

F_net = m1 * g

For the 6.50 kg crate, the net force acting on it is the force of gravity pulling it downward minus the tension in the string:

F_net = m2 * g - T

Since the crates are connected by a light string passing over a frictionless pulley, the acceleration of both crates is the same. Therefore, we can set the expressions for the net force equal to each other:

m1 * g = m2 * g - T

Now, let's solve for T (tension in the string):

T = m2 * g - m1 * g

Plugging in the values we know:

T = 6.50 kg * 9.8 m/s^2 - 10.0 kg * 9.8 m/s^2

Calculating the value:

T ≈ 63.67 N

Therefore, the tension in the string is approximately 63.67 N.