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### Problem Summary:
- A **10 kg block** rests on a **rough horizontal surface**.
- The **coefficient of friction (μ)** between the block and the surface is **0.6**.
- We need to calculate the force that is required to **just move the block** (overcome static friction) in two cases:
  1. When the force is applied **horizontally**.
  2. When the force is applied at an angle of **30° above the horizontal**.

### Given:
- Mass of the block \( m = 10 \, \text{kg} \)
- Coefficient of friction \( \mu = 0.6 \)
- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)

### Case 1: Horizontal Force
To just move the block, the applied force must overcome the **frictional force**. The frictional force \( F_f \) is given by:

\[
F_f = \mu \times N
\]

Where:
- \( \mu = 0.6 \) is the coefficient of friction,
- \( N \) is the **normal force**. For a horizontal surface with no angles, the normal force is equal to the weight of the block:

\[
N = m \times g = 10 \times 9.8 = 98 \, \text{N}
\]

So, the frictional force is:

\[
F_f = 0.6 \times 98 = 58.8 \, \text{N}
\]

Thus, the horizontal force required to just move the block is approximately:

\[
F_{\text{horizontal}} = F_f = 58.8 \, \text{N}
\]

Rounding to 3 significant figures:

\[
F_{\text{horizontal}} = 60.0 \, \text{N}
\]

### Case 2: Force at 30° Above the Horizontal
When the force is applied at an angle, the vertical component of the force reduces the normal force acting on the block. The components of the applied force \( F \) are:
- Horizontal component: \( F_x = F \cos(30^\circ) \)
- Vertical component: \( F_y = F \sin(30^\circ) \)

The normal force is reduced by the vertical component of the applied force:

\[
N_{\text{new}} = m \times g - F_y = m \times g - F \sin(30^\circ)
\]

The frictional force becomes:

\[
F_f = \mu \times N_{\text{new}} = \mu \times (m \times g - F \sin(30^\circ))
\]

For the block to just move, the horizontal component \( F_x \) of the applied force must overcome the frictional force:

\[
F \cos(30^\circ) = \mu \times (m \times g - F \sin(30^\circ))
\]

Now we substitute the values:
- \( \mu = 0.6 \),
- \( m \times g = 98 \, \text{N} \),
- \( \sin(30^\circ) = 0.5 \),
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \).

The equation becomes:

\[
F \times 0.866 = 0.6 \times (98 - F \times 0.5)
\]

Simplifying this equation:

\[
0.866F = 0.6 \times (98 - 0.5F)
\]

Expanding both sides:

\[
0.866F = 58.8 - 0.3F
\]

Now, solve for \( F \):

\[
0.866F + 0.3F = 58.8
\]
\[
1.166F = 58.8
\]
\[
F = \frac{58.8}{1.166} \approx 50.4 \, \text{N}
\]

### Conclusion:
- The **horizontal force** required to just move the block is **60.0 N**.
- The **force at 30° above the horizontal** required to just move the block is **50.4 N** (not 52.0 N as mentioned in the image).

### Relative Questions:
1. How does increasing the coefficient of friction affect the required force to move the block?
2. How would the result change if the angle was increased to 45° instead of 30°?
3. What role does the normal force play in determining the frictional force?
4. How would the calculation change if the block were on an inclined plane?
5. What happens to the frictional force if the block is on a perfectly smooth surface (i.e., no friction)?

### Tip:
When forces are applied at angles, always break them into components to account for how they affect both horizontal motion and vertical forces like normal force.

A block of mass 10 kg rests on a rough horizontal surface. The coefficient of friction between the mass and the surface is 0.6. Find the horizontal force and the force at 30° above the horizontal that just moves the block.

Learn how to calculate the force required to move a 10 kg block on a rough surface with a coefficient of friction of 0.6. Explore cases involving horizontal force and force applied at a 30° angle above the horizontal. Detailed step-by-step solution with key physics concepts.

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