Two forces 3N and 4N on a body in directions due North and due East respectively.Calculate their equilibrant.

Answer 1

X = 4N.
Y = 3N.

X + Yi + F3 = 0
4 + 3i + F3 = 0
F3 = -4 - 3i

Tan Ar = Y/X = -3/-4 = 0.75000
r = 36.87o(Q1). = Reference angle.
A = 36.87 + 180 = 216.87o(Q3).

F3 = X/Cos A = -4/Cos216.87 = 5.0 N. @
216.87o.

Note: F3 is = to and opposite of the sum of the two given forces.

Answer 2

The answer is 10.31km/h

Answer 3

Resultant force(F) =square root of(3square +4square).
F=5N.
Equilibriant has the same magnitude as resultant force.Hence the equilibriant is 5N.

Answer 4

How did we get 5N as the force

Answer 5

Ok

Answer 6

I can't find the perfect solution

Answer 7

Two forces 3N and 4N act on a body in the direction due north and east respectively.calculate the equilibrium

Answer 8

Two forces 3n and 4n act in a direction due north and due East respectively calculate their equilibrium