To solve this problem, we need to consider the forces acting on both the block and the wedge and apply the laws of forces and motion. Here's the step-by-step approach to figuring out the maximum and minimum values of the horizontal force F for which the block does not slip.
Step 1: Identify the forces acting on the block and the wedge.
- For the block: the weight mg (mg = mass × gravitational acceleration), the normal force N perpendicular to the incline, and the static friction force fs parallel to the incline.
- For the wedge: the weight Mg (Mg = mass × gravitational acceleration), the normal force N' perpendicular to the incline (acting from the block), and the static friction force fs' parallel to the incline (acting from the block).
Step 2: Resolve the weight of the block into components.
The weight of the block has two components: one parallel to the incline (mg sinθ) and one perpendicular to the incline (mg cosθ), where θ is the angle of the incline (35° in this case).
Step 3: Apply Newton's second law of motion.
Consider the forces in the horizontal and vertical directions separately for the block and the wedge.
- Horizontal direction: The sum of the forces must be zero (since the block does not slip).
- Vertical direction: The sum of the forces must be zero (since there is no vertical acceleration).
Step 4: Express the forces in terms of their respective equations.
Let's denote fs and fs' as the static friction forces on the block and wedge, respectively.
For the block:
- In the horizontal direction: fs = F, where F is the horizontal force applied.
- In the vertical direction: N = mg cosθ.
For the wedge:
- In the horizontal direction: fs' = F.
- In the vertical direction: N' = mg cosθ.
Step 5: Apply the conditions for static equilibrium.
To prevent slipping, the maximum and minimum values of the force F occur when the static friction force fs is at its maximum and minimum, respectively. So, we can use the formula for the maximum static friction force: fs max = µs N.
Step 6: Substitute the forces' equations into the conditions for static equilibrium.
- For the maximum value of F:
Since fs max = µs N, we can substitute fs with µs N in the horizontal direction equation for the block and wedge: µs N = F.
Substituting N with mg cosθ, we get µs mg cosθ = F.
- For the minimum value of F:
fs min = -µs N (since the force vector direction is changed), so we substitute fs with -µs N in the horizontal direction equation for the block and wedge: -µs N = F.
Substituting N with mg cosθ, we get -µs mg cosθ = F.
Step 7: Calculate the maximum and minimum values of F.
- Substituting the given values into the formula for the maximum value of F:
µs = 0.30, m = 0.40 kg, g = 9.8 m/s^2, and θ = 35°.
F max = (0.30)(0.40 kg)(9.8 m/s^2)(cos 35°).
- Substituting the given values into the formula for the minimum value of F:
µs = 0.10, m = 0.40 kg, g = 9.8 m/s^2, and θ = 35°.
F min = -(0.10)(0.40 kg)(9.8 m/s^2)(cos 35°).
Step 8: Calculate the final numerical values for F max and F min.
By substituting the values into the equations in Step 7 and evaluating them, you can calculate the maximum and minimum values of F.
Remember, these steps provide a general approach to solving the problem. Make sure to double-check the calculations and units for accuracy.