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PhysicsLaws of Motion
The magnitude and direction of the acceleration produced in a body of mass 5 kg when two mutually perpendicular forces 8 N and 6 N act on it, are respectively :
Options
1
20 m s⁻²; tan⁻¹(4/3) with 8N force
2
2 m s⁻²; tan⁻¹(3/4) with 6N force
3
2 m s⁻²; tan⁻¹(4/3) with 8N force
4
2 m s⁻²; tan⁻¹(3/4) with 8N force
Correct Answer
Option 4 : 2 m s⁻²; tan⁻¹(3/4) with 8N force
Step-by-Step Solution
1

Find the resultant force:

The two forces are perpendicular (90° apart). Using Pythagoras:

$$F_{net} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ N}$$

2

Find the magnitude of acceleration:

$$a = \frac{F_{net}}{m} = \frac{10}{5} = \mathbf{2 \text{ m/s}^2}$$

3

Find the direction:

The angle θ with the 8N force:

$$\tan\theta = \frac{6}{8} = \frac{3}{4}$$

$$\theta = \tan^{-1}\!\left(\frac{3}{4}\right) \approx 37° \text{ with the 8N force}$$

✓

Note the 3-4-5 triple: 6-8-10 is just 2×(3-4-5). This is a standard Pythagorean triple that appears very frequently in NEET problems. Recognise it instantly!

Theory: Vectors, Forces & Newton's Laws
1. Newton's Second Law — Vector Form

Newton's Second Law states F⃗_net = ma⃗. The acceleration vector is always in the same direction as the net force vector. When multiple forces act on a body, we must find their vector sum (resultant) first, then divide by mass. The direction of acceleration is determined by the direction of net force, not individual forces.

2. Resultant of Two Perpendicular Forces

\(R = \sqrt{F_1^2 + F_2^2}\) (magnitude)

\(\theta = \tan^{-1}\!\left(\dfrac{F_2}{F_1}\right)\) with \(F_1\)

For perpendicular forces, the parallelogram law reduces to the Pythagorean theorem since sin90° = 1 and cos90° = 0. The resultant always lies between the two forces, making angle θ with the larger force.

3. Pythagorean Triples for NEET

📌 3-4-5: Forces 3N, 4N → R = 5N (very common)

📌 6-8-10: Forces 6N, 8N → R = 10N (this problem)

📌 5-12-13: Forces 5N, 12N → R = 13N

📌 8-15-17: Forces 8N, 15N → R = 17N

📌 Always check if forces form a Pythagorean triple — avoids long calculations!

4. Resolution of Forces

Any force F at angle θ can be resolved into two perpendicular components: F_x = Fcosθ (horizontal) and F_y = Fsinθ (vertical). To find the resultant of multiple forces: sum all x-components (ΣF_x) and all y-components (ΣF_y), then R = √(ΣF_x)² + (ΣF_y)², direction = tan⁻¹(ΣF_y/ΣF_x). This component method works for any number of forces at any angles.

5. Key Angles: 37° and 53°

📌 For 3-4-5 triangle: sin37° = 3/5, cos37° = 4/5, tan37° = 3/4

📌 For 3-4-5 triangle: sin53° = 4/5, cos53° = 3/5, tan53° = 4/3

📌 These angles appear constantly in NEET — memorise them!

📌 Here: θ = tan⁻¹(3/4) ≈ 37° with the 8N force

6. Resultant Force for Non-Perpendicular Forces

For two forces F₁ and F₂ at angle α between them:

\(R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\alpha}\)

Special cases: α = 0° (parallel) → R = F₁ + F₂ (maximum). α = 180° (antiparallel) → R = |F₁ − F₂| (minimum). α = 90° (perpendicular) → R = √(F₁² + F₂²). This problem is the perpendicular case.

Frequently Asked Questions
1. Why is the resultant force 10N and not 14N? ⌄
Forces are vectors, not scalars. We cannot simply add 8 + 6 = 14. Since the forces are perpendicular, we use Pythagoras: R = √(8² + 6²) = √100 = 10 N. Simple addition (14N) applies only when forces act in the same direction.
2. Why is tan⁻¹(3/4) and not tan⁻¹(4/3)? ⌄
The angle with the 8N force uses the opposite (6N) over adjacent (8N): tanθ = 6/8 = 3/4. If the angle were with the 6N force, it would be tan(α) = 8/6 = 4/3. The question asks the angle with the 8N force, so tan⁻¹(3/4).
3. What is tan⁻¹(3/4) in degrees? ⌄
tan⁻¹(3/4) ≈ 36.87° ≈ 37°. Memorise: sin37° ≈ 3/5 = 0.6, cos37° ≈ 4/5 = 0.8, tan37° ≈ 3/4 = 0.75. This 3-4-5 angle appears in almost every NEET question on force components.
4. Can acceleration be in a different direction than net force? ⌄
No. Newton's second law F = ma is a vector equation. Acceleration is always in the exact same direction as the net force. If someone tells you acceleration and force are in different directions, something is wrong with the problem setup.
5. What if the forces were parallel (same direction)? ⌄
If 8N and 6N were in the same direction, R = 14N, a = 14/5 = 2.8 m/s². If they were in opposite directions, R = 2N, a = 0.4 m/s². Perpendicular forces give R = 10N — in between the two extremes.
6. What is the 6-8-10 Pythagorean relationship? ⌄
6-8-10 is the 3-4-5 triple scaled by factor 2. Pythagorean triple means a² + b² = c²: 6² + 8² = 36 + 64 = 100 = 10². Recognising this avoids the need to calculate √100.
7. How would you solve if three forces acted on the body? ⌄
Resolve all three forces into x and y components. Sum all x-components: ΣFx. Sum all y-components: ΣFy. Net force R = √(ΣFx² + ΣFy²). Acceleration = R/m. Direction = tan⁻¹(ΣFy/ΣFx).
8. What is the angle between the resultant and the 6N force? ⌄
If θ = tan⁻¹(3/4) ≈ 37° with the 8N force, then the angle with the 6N force = 90° − 37° = 53°. Verify: tan53° = 4/3, and tan(angle with 6N force) = 8/6 = 4/3 ✓.
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