A box of mass 2 kg is placed on a inclined plane that makes 30\(^\circ\) with the horizontal. The coefficient of friction between the box and inclined plane is 0.2. A force F is applied on the box perpendicular to the incline to prevent the box from sliding down. The minimum value of F is (acceleration due to gravity = 10 m s\(^{-2}\))
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Draw a free-body diagram with all forces.
Resolve forces parallel and perpendicular to the incline.
Normal force N is perpendicular to the surface. If an external force F presses the object into the incline, N increases.
Maximum static friction \(f_{s,max} = \mu_s N\).
For equilibrium on the verge of sliding down: Downward component of weight = Upward frictional force.
Mass of the box \(m = 2\) kg. Angle of inclination \(\theta = 30^\circ\).
Coefficient of friction \(\mu = 0.2\). Acceleration due to gravity \(g = 10 \text{ ms}^{-2}\).
Force F is applied perpendicular to the incline, pressing the box into the incline.
The box is to be prevented from sliding down. This means the frictional force will act upwards along the incline. For the minimum F, we assume the box is on the verge of sliding down, so static friction is at its maximum value, \(f_s_max = \mu N\), where N is the normal reaction force.
Forces acting on the box:
1. Weight \(mg\) acting vertically downwards.
Component parallel to incline, downwards: \(mg\sin\theta\).
Component perpendicular to incline, into the incline: \(mg\cos\theta\).
2. Applied force F, perpendicular to the incline, pressing into the incline.
3. Normal reaction force N, perpendicular to the incline, outwards from the incline.
4. Frictional force \(f_s\), parallel to the incline, upwards (opposing impending motion).
Equilibrium of forces perpendicular to the incline:
\(N - mg\cos\theta - F = 0\)
\(N = mg\cos\theta + F\).
Equilibrium of forces parallel to the incline (verge of sliding down):
\(mg\sin\theta - f_s = 0\)
\(mg\sin\theta = f_s\).
Since it's minimum F, friction is maximum: \(f_s = f_{s,max} = \mu N\).
So, \(mg\sin\theta = \mu N\).
Substitute \(N = mg\cos\theta + F\):
\(mg\sin\theta = \mu (mg\cos\theta + F)\).
We need to find F.
\(mg\sin\theta = \mu mg\cos\theta + \mu F\)
\(\mu F = mg\sin\theta - \mu mg\cos\theta\)
\(F = \frac{mg(\sin\theta - \mu\cos\theta)}{\mu}\).
This formula is for the case where F is needed to *initiate* upward motion or if applied differently. Let's re-evaluate the forces.
F is applied *perpendicular* to the incline.
Normal force \(N = mg\cos\theta + F\). (Correct)
Force pulling down the incline = \(mg\sin\theta\).
Maximum static friction opposing this = \(f_s = \mu N = \mu (mg\cos\theta + F)\).
To prevent sliding down, we need \(f_s \ge mg\sin\theta\).
For minimum F, we consider the limiting case: \(f_s = mg\sin\theta\).
So, \( \mu (mg\cos\theta + F) = mg\sin\theta \).
\( \mu mg\cos\theta + \mu F = mg\sin\theta \).
\( \mu F = mg\sin\theta - \mu mg\cos\theta \).
\( F = \frac{mg(\sin\theta - \mu\cos\theta)}{\mu} = mg \left( \frac{\sin\theta}{\mu} - \cos\theta \right) \).
Let's plug in values: \(m=2, g=10, \theta=30^\circ, \mu=0.2\).
\(\sin 30^\circ = 1/2 = 0.5\).
\(\cos 30^\circ = \sqrt{3}/2 \approx 0.866\).
\(F = (2)(10) \left( \frac{0.5}{0.2} - 0.866 \right) = 20 (2.5 - 0.866)\)
\(F = 20 (1.634) = 32.68 \text{ N} \approx 32.7 \text{ N}\).
This matches option (c).
Check for condition: For friction to exist, \(mg\sin\theta - \mu mg\cos\theta\) must be positive for F to be positive (pushing).
If \(mg\sin\theta \le \mu mg\cos\theta\) (i.e., \(\tan\theta \le \mu\)), the block would not slide down even with F=0.
Here \(\tan 30^\circ = 1/\sqrt{3} \approx 0.577\). \(\mu=0.2\).
Since \(\tan 30^\circ>\mu\) (0.577>0.2), the block will slide down if F is not applied or is too small. So a positive F is needed.
The calculation \(F = mg (\frac{\sin\theta}{\mu} - \cos\theta)\) assumes F is applied to increase normal force. This derivation is correct.
\[ \boxed{32.7 \text{ N}} \]
