Two packing crates of masses 10.89 kg and 2.59 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The 2.59 kg crate lies on a smooth incline of angle 40.0°.
a) Find the acceleration of the 2.59 kg crate
B) find tension in the string
F1 = m*g = 10.89kg * 9.8N/kg = 106.7 N.
F2 = 2.59kg * 9.8N/kg = 25.4 N.@ 40o.
Fp = 25.4*sin40 = 16.3 N. = Force paral-
lel to incline.
a. Fn = F1 - Fp = m*a.
106.7-16.3 = 2.59*a
90.4 = 2.59*a
a = 90.4 m/s^2.
To answer these questions, we can apply Newton's second law of motion:
1) Find the acceleration of the 2.59 kg crate:
Since the pulley is frictionless, the tension in the string on each side of the pulley is the same. Let's call this tension T.
The weight of the 10.89 kg crate can be calculated using the formula: weight = mass * acceleration due to gravity. Therefore, the weight of the 10.89 kg crate is (10.89 kg) * (9.8 m/s^2) = 106.722 N.
Now, let's consider the forces acting on the 2.59 kg crate on the incline:
- The weight acts vertically downward, which can be calculated using the formula: weight = mass * acceleration due to gravity. Therefore, the weight of the 2.59 kg crate is (2.59 kg) * (9.8 m/s^2) = 25.322 N.
- The normal force acts perpendicular to the incline and cancels out the y-component of the crate's weight. This force can be calculated using the formula: normal force = weight * cos(angle of incline). Therefore, the normal force is (25.322 N) * cos(40°) = 19.425 N.
- The force of friction acts parallel to the incline and opposes the direction of motion. The force of friction can be calculated using the formula: force of friction = coefficient of friction * normal force. Since the incline is stated to be smooth, there is no friction present (i.e., coefficient of friction = 0).
Now, let's consider the forces acting on the 10.89 kg crate:
- The weight acts vertically downward, which is (10.89 kg) * (9.8 m/s^2) = 106.722 N.
- The tension in the string acts upwards and can be represented as T. As mentioned earlier, this tension is the same on each side of the pulley.
Using Newton's second law, we can write the following equation for the 2.59 kg crate:
sum of forces = mass * acceleration
In the x-direction (along the incline), this equation becomes:
T - force of friction = mass * acceleration
Since the incline is stated to be smooth (no friction), the force of friction is 0. Therefore, the equation simplifies to:
T = mass * acceleration
Using Newton's second law for the 10.89 kg crate, we have:
sum of forces = mass * acceleration
In the y-direction (perpendicular to the incline), this equation becomes:
Weight - T = mass * acceleration
Substituting the weight for the 10.89 kg crate, we get:
106.722 N - T = (10.89 kg) * acceleration
Since the two crates are connected by a light string that passes over the pulley, the acceleration of one crate will be the same as the other. Therefore, we can equate the two equations:
mass * acceleration = (10.89 kg) * acceleration
mass * acceleration = 106.722 N - T
Rearranging the equations, we get:
T = (10.89 kg) * acceleration
106.722 N - T = (10.89 kg) * acceleration
Now, we have two equations with two unknowns (T and acceleration). We can solve this system of equations to find the values.
2) Find the tension in the string:
Substitute the value of T from the first equation into the second equation:
106.722 N - (10.89 kg) * acceleration = (10.89 kg) * acceleration
Now, we can solve for acceleration:
106.722 N = 2 * (10.89 kg) * acceleration
acceleration = 106.722 N / (2 * 10.89 kg) = 4.913 m/s^2
Finally, substitute the value of acceleration into the equation for T to find the tension:
T = (10.89 kg) * acceleration = (10.89 kg) * (4.913 m/s^2) = 53.414 N
Hence, the acceleration of the 2.59 kg crate is 4.913 m/s^2, and the tension in the string is 53.414 N.