the sum of two foces acting at point is 16N if resultant force is 8N and direction is perpendicular to minimum forcethen the force are
Answers
Answer:
Smaller force is 6N and larger force is 10N
Explanation:
Let the smaller of the two forces is of magnitude be
F
N.
Hence the larger one will be
(
16
−
F
)
N. As the resultant
8
N is perpendicular to
F
we can write.
F
2
+
8
2
=
(
16
−
F
)
2
⇒
(
16
−
F
)
2
−
F
2
=
64
⇒
16
(
16
−
2
F
)
=
64
⇒
256
−
32
F
=
64
⇒
F
=
256
−
64
32
=
6
N
So the smaller force is
6
N
and the larger force is
10
N
Hope that helps you,my buddy.Please mark me as the brainliest.
hey meet your answers
Let the smaller of the two forces is of magnitude be
F N.
Hence the larger one will be
(16−F)
N. As the resultant 8
N is perpendicular to F we can write.
F2+82=(16−F)2
⇒(16−F)2−F2=64
⇒16(16−2F)=64
⇒256−32F=64
⇒F=256−6432=6N
So the smaller force is
6
N
and the larger force is
10 N
I hope it helps you please make me brainliest