A group of Mechanical Engineering students are attempting to pull a large crate of mass M=175.0kg on a level driveway. The coefficient of static friction ms = 0.610. They use a rope of dubious quality to pull the crate as indicated in the figure. At what angle with respect to the ground should they pull to minimize the chance that the rope will break while they attempt to get the crate to start sliding (in degrees)?
To determine the angle at which the Mechanical Engineering students should pull the crate to minimize the chances of the rope breaking, we need to analyze the forces acting on the crate.
1. Identify the forces acting on the crate:
- Weight (mg): The force due to the mass of the crate acting vertically downward.
- Normal force (N): The force exerted by the ground perpendicular to the surface of the crate.
- Tension force (T): The force exerted by the rope pulling the crate horizontally.
- Friction force (F): The force opposing motion between the crate and the ground.
2. Determine the forces' magnitudes:
- Weight (mg) = 175.0 kg * 9.8 m/s^2 = 1715 N (approximately)
- Normal force (N) = Weight = 1715 N (since it acts perpendicular to the surface)
3. Analyze the friction force:
The maximum static friction force (Fs) can be calculated using the coefficient of static friction (μs) and the normal force (N):
- Fs = μs * N
- Fs = 0.610 * 1715 N = 1046.15 N (approximately)
4. Resolve the tension force (T):
Since we want to minimize the chance of the rope breaking, we need to ensure that the tension force does not exceed the maximum static friction force.
- T <= Fs
5. Resolve the tension force components:
The tension force can be resolved into horizontal and vertical components.
- T_horizontal = T * cosθ
- T_vertical = T * sinθ
(θ represents the angle at which they should pull the crate.)
6. Determine the maximum value for the horizontal tension force component (T_horizontal):
Since T_horizontal needs to be equal to or less than the maximum static friction force, we have:
- T_horizontal <= Fs
- T * cosθ <= Fs
- T <= Fs / cosθ
7. Substitute the maximum value of T into T_vertical:
Using the equation:
- T_horizontal <= Fs / cosθ
- T_vertical = T * sinθ
We can substitute the value of T from the previous step into T_vertical:
- T_vertical = (Fs / cosθ) * sinθ
8. Simplify the equation:
- T_vertical = (Fs * sinθ) / cosθ
- T_vertical = Fs * tanθ
9. Determine the maximum value for the vertical tension force component (T_vertical):
The maximum value for T_vertical should not exceed the crate's weight. Therefore:
- T_vertical <= Weight
- Fs * tanθ <= mg
10. Calculate the angle (θ):
To determine the minimum angle at which they should pull the crate, we need to find the value of θ that satisfies the inequality:
- Fs * tanθ <= mg
- 1046.15 N * tanθ <= 175.0 kg * 9.8 m/s^2
- tanθ <= 16.4543 (approximately)
11. Find the angle (θ) using an inverse tangent (tan^-1) function:
- θ = tan^-1(16.4543)
- θ ≈ 87.2 degrees
Therefore, the Mechanical Engineering students should pull the crate at an angle of approximately 87.2 degrees with respect to the ground to minimize the chance of the rope breaking.