Two packing crates of masses m1 = 10.0 kg and m2 = 4.10 kg are connected by a light string that passes over a frictionless pulley as in the figure. The 4.10-kg crate lies on a smooth incline of angle 33.0°.
(a) Find the acceleration of the 4.10-kg crate.
(b) Find the tension in the string.
To find the acceleration of the 4.10 kg crate and the tension in the string, we can use Newton's second law and analyze the forces acting on the system.
First, let's calculate the weight of each crate. The weight of an object can be found using the formula:
Weight = mass x acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s^2.
For the 10.0 kg crate:
Weight1 = 10.0 kg x 9.8 m/s^2
For the 4.10 kg crate:
Weight2 = 4.10 kg x 9.8 m/s^2
Next, let's analyze the forces acting on the system. We have the following forces:
For the 10.0 kg crate:
- Tension in the string, pulling it up the incline (T1)
- Weight of the crate, pulling it down the incline (Weight1)
For the 4.10 kg crate:
- Tension in the string, pulling it down the incline (T2)
- Weight of the crate, pulling it down the incline (Weight2)
Now, let's decompose the weight of the 4.10 kg crate into two components:
- Weight component parallel to the incline (Weight2_parallel): Weight2 x sin(33.0°)
- Weight component perpendicular to the incline (Weight2_perpendicular): Weight2 x cos(33.0°)
Since the crates are connected by a light string, their accelerations will be the same. Let's denote this acceleration as 'a'.
Using Newton's second law, we can write the equation for each crate:
For the 10.0 kg crate:
T1 - Weight1 = 10.0 kg x a
For the 4.10 kg crate:
T2 + Weight2_parallel = 4.10 kg x a
Now, let's eliminate 'T1' and 'T2' from these equations by solving them simultaneously.
From the first equation, we can express 'T1' as:
T1 = Weight1 + 10.0 kg x a
Substituting 'T1' into the second equation:
(Weight1 + 10.0 kg x a) + Weight2_parallel = 4.10 kg x a
Now, substitute the values we have calculated earlier:
(Weight1 + 10.0 kg x a) + (Weight2 x sin(33.0°)) = 4.10 kg x a
Simplify the equation:
Weight1 + (10.0 kg + Weight2 x sin(33.0°)) = 4.10 kg x a
Solve for 'a':
a = (Weight1 + Weight2 x sin(33.0°)) / (10.0 kg + 4.10 kg)
Now that we have the acceleration 'a', we can calculate the tension in the string 'T2' using the second equation:
T2 + Weight2_parallel = 4.10 kg x a
T2 = 4.10 kg x a - Weight2_parallel
Substitute the calculated values:
T2 = 4.10 kg x a - (Weight2 x sin(33.0°))
Now, you can substitute the values of Weight1, Weight2, and the angle into the equations to find the values of 'a' and 'T2'.