Two packing crates of masses 7.88 kg and 4.62 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The 4.62 kg crate lies on a smooth incline of angle 40.0°.
a) Find the acceleration of the 4.62 kg crate
m2•a=T - m2•g•sinα => T= m2•a + m2•g•sinα
m1•a=m1•g – T= m1•g - m2•a - m2•g•sinα.
a=g•(m1-m2•sinα)/(m1+m2) = 9.8•(7.88-4.62• sin40°)/(7.88+4.62) =
=.... m/s²,
T= m1(g-a)= 7.88•(9.8- a) = ... N.
To find the acceleration of the 4.62 kg crate, we need to consider the forces acting on it. The forces acting on the crate are its weight, the tension in the string, and the normal force.
1. Weight: The weight of the crate is given by the equation W = m * g, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2). The weight of the 4.62 kg crate is W1 = 4.62 kg * 9.8 m/s^2.
2. Tension: The tension in the string can be calculated using the tension in one side of the string. Since the string is light and frictionless, the tension is the same on both sides. We'll call the tension T.
3. Normal Force: The normal force is the force exerted by the incline perpendicular to the surface. It can be calculated using the equation N = m * g * cos(theta), where theta is the angle of the incline. The mass for calculation is the component of the crate's mass perpendicular to the incline, which is m1 * cos(theta), where m1 is the mass of the crate.
Now, we can write the equation of motion for the crate:
m1 * a = T - m1 * g * sin(theta)
T - m1 * g * sin(theta) = m1 * a
Substituting the values:
4.62 kg * a = T - (4.62 kg * 9.8 m/s^2 * sin(40.0°))
To solve for the acceleration, we need the value of tension (T). The tension is equal to the weight of the other crate (7.88 kg * 9.8 m/s^2) due to their connection through the light string:
T = 7.88 kg * 9.8 m/s^2
Substituting this value back into the equation:
4.62 kg * a = (7.88 kg * 9.8 m/s^2) - (4.62 kg * 9.8 m/s^2 * sin(40.0°))
Now solve for acceleration:
a = [(7.88 kg * 9.8 m/s^2) - (4.62 kg * 9.8 m/s^2 * sin(40.0°))] / 4.62 kg
To find the acceleration of the 4.62 kg crate, we need to consider the forces acting on it.
First, let's identify the forces involved:
1. The weight of the 4.62 kg crate: This force acts vertically downward and can be calculated using the formula F = mg, where m is the mass (4.62 kg) and g is the acceleration due to gravity (9.8 m/s^2).
2. The tension in the string: This force acts horizontally and is the same for both crates. It can be calculated using the formula T = m2a, where m2 is the mass of the 7.88 kg crate and a is the acceleration.
3. The normal force: This force acts perpendicular to the incline and counteracts the component of the weight force acting along the incline. However, since the incline is smooth (frictionless), there is no friction force.
4. The gravitational force component along the incline: This force acts parallel to the incline and is given by mg sinθ, where θ is the angle of the incline (40.0°).
The net force acting on the 4.62 kg crate is the difference between the gravitational force component along the incline (mg sinθ) and the tension in the string (T).
The net force can be determined using the equation:
Net force = mg sinθ - T
Now, since the acceleration of both crates is the same and the masses are connected by a light string, the tension in the string will be the same as the net force on the 4.62 kg crate.
So, we can rewrite the equation as:
T = mg sinθ - T
Simplifying this equation, we get:
2T = mg sinθ
Substituting the values into the equation, we have:
2T = (4.62 kg)(9.8 m/s^2) sin(40.0°)
Now, solve for T:
T = [(4.62 kg)(9.8 m/s^2) sin(40.0°)] / 2
Calculate T:
T ≈ 15.92 N
Since the tension in the string is the same as the net force acting on the 4.62 kg crate, we can use this value of T to find the acceleration:
T = m2a
15.92 N = (4.62 kg) * a
Solve for a:
a ≈ 3.45 m/s^2
Therefore, the acceleration of the 4.62 kg crate is approximately 3.45 m/s^2.