Let's break down the problem and understand how to solve it.

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.