if two forces 10N and 20N are inclined at angle 60degree to each other. Find the resultant force.
if two forces 10N and 20N are inclined at angle 60degree to each other. Find the resultant force.
R^2= a^2+c^2-2ac cos(180 - THETA) Theta in this case is 60, 180-60=120
R^2 = 10^2 + 20^2 – (2 x 10 x 20 cos 120)
R^2 = 700N
R=√700N
R= √100 * √7 N
R= 10*√7 N or 10√7N
R=f1+f2=10+20(both in square) R=100+400=500.(find the square root)=22.4n.therefore resultant force is 22.4n
R=20^2+10^ -2of10cos[120]=400+100-400[-0.5]=500+200=700
NOT CLEAR ENOUGH
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R²=400 +100 - 400cos120
=500+200
√700
R=26.5N
R²=a²+b²-2ab cos theta
But R?
a=10N
b=20N
Theta=(180-60)=120°
R²=10²+20²-2×10×20 cos 120°
=100+400-400(0.5)
=500+400×0.5
=500+200
=700
R²=√700
=26.5N,,
No. 1
√300
I don't understand please explain better 😞
R=20^2+10^-2 of 10 cos [120]
=400+100-400[-0.5]
=500+200=700
F1 is 10n while f2 is 20n while angle is 60°
r^2=20^2+10^2-2×20×10 cos 120
r^2=500+200
r^2=700
r^2=26.5
i am baba from india i will solve everything
F = 10N.[0o] + 20[60o] = 10 + 20*Cos60 +
i20*sin60 = 10 + 10 + 17.3i = 20 + 17.3i
= 26.5N.[40.9o].