Physics · Laws Of Motion · NEET
No. You can only add force magnitudes directly when the forces point in the SAME straight line. When they are perpendicular (at 90 degrees), you must combine them as vectors: F = sqrt(8^2 + 6^2) = sqrt(64+36) = sqrt(100) = 10 N. Adding to get 14 N is the most common mistake - it ignores that the forces pull in different directions.
Acceleration always points in the same direction as the NET (resultant) force, because a = F_net/m and mass is a positive scalar. So it points along the diagonal, not along the 8 N force or the 6 N force alone. You describe this direction by the angle it makes with one of the forces, using tan(theta) = opposite force / chosen force.
Choose the force you want to measure the angle from (usually the larger one, here 8 N). The angle is theta = tan^-1(other force / chosen force). Measuring from the 8 N force: tan(theta) = 6/8 = 3/4, so theta = tan^-1(3/4) ~ 37 degrees. If you measured from the 6 N force instead you would get tan^-1(8/6) = tan^-1(4/3) ~ 53 degrees. Both describe the same diagonal - just from different reference forces, and they add to 90.
Newton's second law is a vector equation: vector a = vector F / m. Dividing a vector by a positive number (mass) shrinks or grows its length but never turns it. So the acceleration is parallel to the net force always; the mass only decides HOW big the acceleration is (a = 10/5 = 2 m/s^2 here).
Then Pythagoras alone is not enough. For a general angle you use the parallelogram law: F = sqrt(F1^2 + F2^2 + 2*F1*F2*cos(angle between them)). The perpendicular case is special because cos(90) = 0, which kills the last term and leaves the clean F = sqrt(F1^2 + F2^2).
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:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
First find the resultant force F = sqrt(F1^2 + F2^2), then a = F/m. The direction is tan^-1(F2/F1) measured from the F1 force.
Because 8 N and 6 N are perpendicular, they combine to 10 N (not 14 N). Then a = 10/5 = 2 m/s^2. The 2.8 value comes from wrongly adding the forces as if they were in the same direction.
Yes. Acceleration has both magnitude (2 m/s^2) and direction (along the resultant force, tan^-1(3/4) from the 8 N force). NEET expects you to state both.
Yes. Forces 6 N and 8 N with resultant 10 N form a 6-8-10 (i.e. 3-4-5) right triangle, so you can spot the resultant as 10 N instantly without full calculation.
Newton's second law F = ma is a vector law. It applies to the NET force. So you must add the two forces vectorially first, then divide by mass to get the acceleration vector.
