To determine the force required to get the crate moving up the inclined plane, we need to consider several factors:

1. Gravitational force component along the incline: This is the component of the crate's weight that acts parallel to the incline and pulls the crate down.
2. Normal force: This is the component of the crate's weight that acts perpendicular to the incline.
3. Frictional force: The force of static friction opposes the motion and depends on the normal force and the coefficient of static friction.
4. Required force: The force that Tony needs to apply to overcome both the gravitational force pulling the crate down the incline and the static friction.

Step 1: Calculate the weight of the crate

The weight $W$ of the crate is given by: $W = mg$ where:

• $m = 30 \, \text{kg}$ (mass of the crate)
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity)

$W = 30 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 294 \, \text{N}$

Step 2: Calculate the gravitational force component along the incline

The component of the gravitational force acting down the incline is: $F_{\text{gravity\_parallel}} = W \sin(\theta)$ where:

• $\theta = 8^\circ$ (angle of the incline)

$F_{\text{gravity\_parallel}} = 294 \, \text{N} \times \sin(8^\circ) \approx 294 \, \text{N} \times 0.1392 \approx 40.92 \, \text{N}$

Step 3: Calculate the normal force

The normal force $F_N$ is the component of the weight perpendicular to the incline: $F_N = W \cos(\theta)$

$F_N = 294 \, \text{N} \times \cos(8^\circ) \approx 294 \, \text{N} \times 0.9903 \approx 291 \, \text{N}$

Step 4: Calculate the maximum static friction force

The maximum static friction force $F_{\text{friction}}$ is given by: $F_{\text{friction}} = \mu_s F_N$ where:

• $\mu_s = 0.35$ (coefficient of static friction)

$F_{\text{friction}} = 0.35 \times 291 \, \text{N} \approx 101.85 \, \text{N}$

Step 5: Determine the required force to move the crate

The force required to just start moving the crate $F_{\text{required}}$ must overcome both the gravitational force component along the incline and the static friction force: $F_{\text{required}} = F_{\text{gravity\_parallel}} + F_{\text{friction}}$

$F_{\text{required}} = 40.92 \, \text{N} + 101.85 \, \text{N} \approx 142.77 \, \text{N}$

Conclusion

The force required to get the crate moving is approximately 143 N.

Correct Answer: A) 143 N

Would you like more details or have any questions? Here are five related questions to expand on this topic:

1. How would the required force change if the angle of the incline increased?
2. What would happen if the coefficient of static friction was reduced?
3. How would you calculate the force required to keep the crate moving at a constant velocity?
4. What is the role of kinetic friction in this scenario, and how does it compare to static friction?
5. How does the mass of the crate affect the normal force on an inclined plane?

Tip: Remember that static friction is usually greater than kinetic friction, meaning it takes more force to start an object moving than to keep it moving.