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Two forces of f1=500N due east &f2=250N due north find f2-f1?
Most Upvoted Answer
Two forces of f1=500N due east &f2=250N due north find f2-f1?
And f2=300N at an angle of 60 degrees north of east are acting on an object. Find the magnitude and direction of the resultant force.

To find the magnitude of the resultant force, we can use the formula:

R² = f₁² + f₂² + 2f₁f₂cosθ

where R is the magnitude of the resultant force, f₁ and f₂ are the magnitudes of the two forces, and θ is the angle between the two forces.

R² = (500)² + (300)² + 2(500)(300)cos60°
R² = 250000 + 90000 + 300000
R² = 640000
R = √640000
R = 800N

To find the direction of the resultant force, we can use trigonometry. Let's call the angle between the resultant force and the east direction θ.

tanθ = (f₂sin60°) / (f₁ + f₂cos60°)
tanθ = (300sin60°) / (500 + 300cos60°)
tanθ = 0.5774 / 0.5
θ = tan⁻¹(1.1547)
θ = 50.19° north of east

Therefore, the magnitude of the resultant force is 800N and its direction is 50.19° north of east.
Community Answer
Two forces of f1=500N due east &f2=250N due north find f2-f1?
F is the resultant ,
F^2 = F1^2 + F2^2 ...cos 90 = 0
= 62500 + 250000
= 875000
but, F2^2 + F1^2 = (F2 - F1)^2+ 2 F1.F2
SO (F2- F1)^2 = F2^2+ F1^2 - 2F1F2
= 875000 - 250000
= 62.5 × 10^4
i.e. F2- F1 = 790
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