Two packing crates of masses 7.57 kg and 4.51 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The 4.51 kg crate lies on a smooth incline of angle 40.0°.
without the diagram description....
12
To find the tension in the string and the acceleration of the system, we can use Newton's second law and the concept of inclined planes.
1. Start by drawing a free-body diagram for each crate separately.
For the 7.57 kg crate:
- There is a gravitational force (mg) pointing downwards.
- There is a tension force (T) pointing upwards.
For the 4.51 kg crate:
- There is a gravitational force (mg) pointing downwards.
- There is a normal force (N) perpendicular to the incline.
- There is a tension force (T) parallel to the incline and pointing upwards.
- There is a friction force (f) parallel to the incline and pointing downwards.
2. Resolve the gravitational force into its components.
For the 7.57 kg crate:
- The vertical component is mg, pointing downwards.
- The horizontal component is zero because the crate is vertical.
For the 4.51 kg crate:
- The vertical component is mg * cos(40°), pointing downwards.
- The horizontal component is mg * sin(40°), pointing towards the incline.
3. Apply Newton's second law in the vertical direction to find the tension in the string.
For the 7.57 kg crate:
- T - (7.57 kg * 9.8 m/s^2) = 0
- T = 7.57 kg * 9.8 m/s^2 = 74.1866 N
For the 4.51 kg crate:
- T - (4.51 kg * 9.8 m/s^2 * cos(40°)) = 0
- T = 4.51 kg * 9.8 m/s^2 * cos(40°) = 29.5889 N
4. Apply Newton's second law in the horizontal direction to find the acceleration of the system.
For the 4.51 kg crate:
- T - f = m * a
- 29.5889 N - f = 4.51 kg * a
5. Apply the friction force equation for the 4.51 kg crate.
The friction force can be calculated using the equation f = u * N, where u is the coefficient of friction and N is the normal force. Since the crate is on a smooth incline, there is no friction, so f = 0.
6. Substitute the values and solve for the acceleration of the system.
- 29.5889 N - 0 = 4.51 kg * a
- a = (29.5889 N) / (4.51 kg)
- a = 6.551 m/s^2
Therefore, the tension in the string is 74.1866 N and the acceleration of the system is 6.551 m/s^2.