Question

the resultant between 2P and sqrt 2P is sqrt 10 P.
What is the angle between them?
@michele_liano

Answer 1

@Michele_Laino

Answer 2

I think that we have to solve this equation:
\[{\left( {2P} \right)^2} + 2P + 2\left( {2P} \right)\sqrt {2P} \cos \theta = 10P\]

Answer 3

please wait I have made a typo

Answer 4

here it is the right equation:
\[{\left( {2P} \right)^2} + 2{P^2} + 2\left( {2P} \right)\sqrt 2 P\cos \theta = 10{P^2}\]

Answer 5

hows there only one sqrt 2 P in that equation?

Answer 6

since the magnitudes of the involved forces, are:
\(2P\), \(\sqrt 2 P\), and \(\sqrt{10} P\)

Answer 7

sqrt 10 P = 6P^2 + 4Psqrt2Pcostheta?

Answer 8

sorry, we have this equation:
\[{\left( {2P} \right)^2} + 2{P^2} - 2\left( {2P} \right)\sqrt 2 P\cos \theta = 10{P^2}\]

Answer 9

wherein I have applied the Carnot Theorem

Answer 10

so,4P^2 = 4sqrt2P^2costheta?

Answer 11

yes! More precisely, we have:
\[ - 4\sqrt 2 \;{P^2}\cos \theta = 4{P^2}\]

Answer 12

how minus?

Answer 13

sorry I have made a typo:
\[4\sqrt 2 \;{P^2}\cos \theta = 4{P^2}\]

Answer 14

if I divide both sides by \(4 \sqrt 2 P^2\), I get:
\[\cos \theta = \frac{{4{P^2}}}{{4\sqrt 2 \;{P^2}}} = ...?\]
please simplify

Answer 15

1/sqrt2

Answer 16

correct!
so, \[\theta = \arccos \left( {\frac{1}{{\sqrt 2 }}} \right) = ...?\]

Answer 17

30??/

Answer 18

If we look at the tables, or we use a calculator, we find \(\theta= \pi/4\)

Answer 19

or \(\theta=45\) degrees

Answer 20

Thanks ,sir.

Answer 21

:)