4 forces act at a point,1st 500 gm,to south,2nd 50 gm to west,400 gm to north,100 gm to east,?

Question

this is the question what is the magnitude and direction of the resultant pls ans. my homework i need to past it 2morow

Answer 1

let unit vectors along east & west be (+i) and (-i)
let unit vectors along north & south be (+j) and (-j)

along East = Force (east) = Force (e) + Force (w)
F(E) = 100 g (i) - 50 g (i) = 50g (i)

along north = Force (north) = Force (n) + Force (s)
F(N) = 400 g (j) - 500 g (j) = - 100g (j)

Resultant force R = F(e) + F(n) = 50g (i) - 100g (j)
magnitude |R| = sqrt [(50g)^2 + (100g)^2]
|R| = 111.8 g = 111.8*980 = 109564 g-cm/s^2
|R| = 109564 * 10^-3kg* 10^-2 m/s^2 = 1.095 Newton

direction tan (p) = |F(n)| / |F(e)| = - 100 /50 = -2
p = 360 deg - 63.44 = 296.56 edgree counter-clockwise from east
you may call this angle as 63.56 deg East of South

Answer 2

you can start by grouping forces on the same line ( opposing directions -- north with south and east with west)
400 gm north combined with 500 gm south means that the northward is cancelled out by part of the southward force

it may help to designate one direction as positive and one as negative (north = +, south = -) so you have:
+400 - 500 = -100 , so 100gm south
do the same for east and west: (east = +, west = -)
+100 - 50 = +50 , so 50 gm east
now to find the resultant force's direction and magnitude use trigonometry:
50 and 100 are the lengths of two sides of a right triangle since south and east are perpendicular . . .

the magnitude can be found using the pythagorean theorem
a^2 + b^2= c^2 where a and b are the known lengths and c is the hypotenuse and therefore the resultant magnitude

to get the direction simply use trigonometry since tangent is opposite over adjacent, find the inverse tangent (tan^-1) of 50/100 (or 1/2)

this will give you the angle (how many degrees east of south the direction vector points)

you will find that the magnitude is roughly 111.8 or 112
and the direction will be roughly 26.6 degrees east of south

Answer 3

Well, a gram is not a unit of force, but using the numbers,
100E - 50W = 50E
50E + 400N =
tanθ = 50/400 = 1/8 = 0.125
θ ≈ 7.1250° E of N
F = 400/cosθ = 400/0.99228
F ≈ 403.11 @ 7.1250° E of N

Check on magnitude:
10√(5^2 + 40^2) =
10√(25 + 1600) =
10√1625 ≈ 403.1129

Answer 4

Helmut you seem to have forgotten the 500gm weight to the south!