A box of mass m = 8.0 kg is resting on a surface with a coefficient of static friction μs = 0.35
and a coefficient of kinetic friction μk = 0.20. A force F is applied at an angle of 40◦ below
the horizontal. Determine the minimum force F required to start moving the box

To determine the minimum force \( F \) required to start moving the box, we need to account for both the vertical and horizontal components of the applied force, as well as the frictional force that opposes the motion.

### Step 1: Components of the applied force
Given that the force is applied at an angle of \( 40^\circ \) below the horizontal, the components of the applied force \( F \) are:
- Horizontal component: \( F_x = F \cos(40^\circ) \)
- Vertical component: \( F_y = F \sin(40^\circ) \) (acting downwards, increasing the normal force)

### Step 2: Calculate the normal force \( N \)
The normal force \( N \) is affected by both the weight of the box and the vertical component of the applied force.

The weight of the box is given by:
\[
W = mg = 8.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 78.4 \, \text{N}
\]

Since the applied force is acting downward, it increases the normal force. The normal force is:
\[
N = W + F_y = mg + F \sin(40^\circ)
\]
\[
N = 78.4 \, \text{N} + F \sin(40^\circ)
\]

### Step 3: Calculate the frictional force
The maximum static friction force \( f_s \) is given by:
\[
f_s = \mu_s N = \mu_s \left( mg + F \sin(40^\circ) \right)
\]
\[
f_s = 0.35 \left( 78.4 \, \text{N} + F \sin(40^\circ) \right)
\]

### Step 4: Set up the equation for equilibrium
For the box to start moving, the horizontal component of the applied force must overcome the static frictional force. Therefore:
\[
F \cos(40^\circ) = f_s
\]
\[
F \cos(40^\circ) = 0.35 \left( 78.4 \, \text{N} + F \sin(40^\circ) \right)
\]

### Step 5: Solve for \( F \)
Now, solve this equation for \( F \). Distribute the friction term on the right-hand side:
\[
F \cos(40^\circ) = 0.35 \times 78.4 \, \text{N} + 0.35 F \sin(40^\circ)
\]
\[
F \cos(40^\circ) - 0.35 F \sin(40^\circ) = 0.35 \times 78.4 \, \text{N}
\]
\[
F \left( \cos(40^\circ) - 0.35 \sin(40^\circ) \right) = 0.35 \times 78.4
\]

Substitute the values of \( \cos(40^\circ) \approx 0.766 \) and \( \sin(40^\circ) \approx 0.643 \):
\[
F \left( 0.766 - 0.35 \times 0.643 \right) = 0.35 \times 78.4
\]
\[
F \left( 0.766 - 0.225 \right) = 27.44
\]
\[
F \times 0.541 = 27.44
\]
\[
F = \frac{27.44}{0.541} \approx 50.7 \, \text{N}
\]

### Final Answer:
The minimum force \( F \) required to start moving the box is approximately **50.7 N**.

Would you like more details on any step or have further questions?

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Here are 5 related questions:
1. How would the answer change if the coefficient of static friction were higher?
2. What happens to the required force if the angle of the applied force changes?
3. How does kinetic friction affect the box once it starts moving?
4. Can we reduce the force required by applying it at a different angle?
5. How is the normal force affected if the surface is inclined?

**Tip:** Always check the direction of the forces, especially when dealing with angles, as vertical components can influence the normal force significantly.

A box of mass m = 8.0 kg is resting on a surface with a coefficient of static friction μs = 0.35 and a coefficient of kinetic friction μk = 0.20. A force F is applied at an angle of 40° below the horizontal. Determine the minimum force F required to start moving the box.

Determine the minimum force needed to move an 8.0 kg box on a surface with static friction coefficient μs = 0.35. This problem applies concepts of static friction, trigonometric force decomposition, and equilibrium equations. Ideal for Grades 11-12 physics students.

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