Fisika SMA - Hukum Newton (7) - Hukum Newton Pada Bidang Miring (1 sisi miring) (I)

Summary

The video explains how to apply Newton’s laws to objects on a single inclined plane. The main strategy is to draw a free-body diagram, choose axes aligned with the slope, resolve gravity into components, and then apply Newton’s second law.

General method

1. Draw all forces acting on the object:
   • Weight (mg), vertically downward.
   • Normal force (N), perpendicular to the plane.
   • Friction, if the plane is rough; it opposes the motion or intended motion.
   • Any applied force, if present.
2. Choose the (x)-axis parallel to the plane, in the direction of motion, and the (y)-axis perpendicular to it.
3. Resolve the weight into components:
   • Along the plane: (mg\sin\theta).
   • Perpendicular to the plane: (mg\cos\theta).
4. Apply Newton’s second law along the slope to find the acceleration.
5. Use the perpendicular direction to find the normal force. Since the object does not accelerate away from or into the plane, the net force perpendicular to it is zero.
6. For a rough plane, calculate friction using (f=\mu N), then include it in the force equation with a sign determined by its direction.

Worked examples

• Smooth plane, object moving downhill at (37^\circ): With no friction, the acceleration is (a=g\sin37^\circ). Using (g=10\text{ m/s}^2) and (\sin37^\circ=0.6), the acceleration is (6\text{ m/s}^2) downhill.
• Rough plane, object moving downhill at (37^\circ), with (\mu=0.2): Friction acts uphill, so (a=g(\sin37^\circ-\mu\cos37^\circ)). Using (\cos37^\circ=0.8), the acceleration is (4.4\text{ m/s}^2) downhill. This illustrates that friction reduces the downhill acceleration.
• Smooth (45^\circ) plane, object pushed uphill by a (12\text{ N}) force: The example uses a mass of approximately (\sqrt2\text{ kg}). The downhill component of gravity is (mg\sin45^\circ=10\text{ N}), leaving a net force of (2\text{ N}) uphill. The acceleration is (\sqrt2\text{ m/s}^2), or about (1.41\text{ m/s}^2) uphill.
• Rough (45^\circ) plane with the same push and (\mu=0.1): Friction acts downhill, opposing the uphill motion. The example’s normal force is (10\text{ N}), so friction is (1\text{ N}). After accounting for gravity’s (10\text{ N}) downhill component and friction, the net force is (1\text{ N}) uphill. The acceleration is (\sqrt2/2\text{ m/s}^2), or about (0.71\text{ m/s}^2) uphill.

Speakers or sources featured

One instructor/presenter from the Le GuruLes tutoring channel is featured. No other speakers or external sources are mentioned.