To solve this problem, we need to find the horizontal force $F$ that will cause the block to start moving up the incline.

Given dataTo solve this problem, we need to find the horizontal force $F$ that will cause the block to start moving up the incline.

Given

• Mass of the block, $m = 4.00 \, \text{kg}$
• Angle of the incline, $\theta = 30.0^\circ$
• Coefficient of static friction, $\mu_s = 0.700$
• Gravitational acceleration, $g = 9.8 \, \text{m/s}^2$

Steps to Solve the Problem:

1. Calculate the gravitational force components:

   • The weight of the block $W = mg = 4.00 \times 9.8 = 39.2 \, \text{N}$.
   • The component of weight parallel to the incline is $W_{\parallel} = W \sin \theta = 39.2 \sin 30^\circ = 19.6 \, \text{N}$.
   • The component of weight perpendicular to the incline is $W_{\perp} = W \cos \theta = 39.2 \cos 30^\circ \approx 33.94 \, \text{N}$.

2. Calculate the maximum static friction force:

   • The normal force $N$ is affected by both $W_{\perp}$ and the horizontal force $F$ we are applying.
   • Initially, without $F$, the normal force $N_0 = W_{\perp} = 33.94 \, \text{N}$.
   • The maximum static friction force is given by $f_s = \mu_s N$.

3. Determine the effect of $F$ on the normal force:

   • The horizontal force $F$ will have a component that contributes to the normal force on the incline.
   • The component of $F$ perpendicular to the incline is $F \sin \theta$.
   • The new normal force $N = W_{\perp} + F \sin \theta$.

4. Set up the equation for force balance along the incline:

   • For the block to start moving up the incline, the sum of forces along the incline must overcome both the gravitational component down the incline and the static friction.
   • The force needed to overcome gravity and friction is: $F \cos \theta = W_{\parallel} + f_s$
   • Substitute $f_s = \mu_s N = \mu_s (W_{\perp} + F \sin \theta)$: $F \cos \theta = W_{\parallel} + \mu_s (W_{\perp} + F \sin \theta)$
   • Rearrange to solve for $F$: $F (\cos \theta - \mu_s \sin \theta) = W_{\parallel} + \mu_s W_{\perp}$ $F = \frac{W_{\parallel} + \mu_s W_{\perp}}{\cos \theta - \mu_s \sin \theta}$

5. Plug in the values: $F = \frac{19.6 + (0.700 \times 33.94)}{\cos 30^\circ - (0.700 \times \sin 30^\circ)}$ Calculating each part:

   • $\mu_s W_{\perp} = 0.700 \times 33.94 \approx 23.758$
   • $W_{\parallel} + \mu_s W_{\perp} = 19.6 + 23.758 = 43.358$
   • $\cos 30^\circ \approx 0.866$
   • $\mu_s \sin 30^\circ = 0.700 \times 0.5 = 0.35$
   • $\cos 30^\circ - \mu_s \sin 30^\circ \approx 0.866 - 0.35 = 0.516$

   So: $F = \frac{43.358}{0.516} \approx 84.0 \, \text{N}$

Answer:

The magnitude of the horizontal force $F$ that must act on the block to start it moving up the incline is 84.0 N. Thus, the correct answer is (B) 84.0 N.

Would you like more details on any part of this solution?

Related Questions:

1. How would the required force change if the incline angle were increased?
2. What would be the effect of reducing the coefficient of static friction?
3. How is the normal force calculated on an inclined plane without friction?
4. How would this problem change if the block were moving rather than static?
5. Can we solve similar problems using energy methods instead of forces?

Tip:

When dealing with inclined planes and forces, always resolve forces parallel and perpendicular to the incline, as it simplifies calculations.