Question

the resultant of 2 forces at right angle is 17N . if the maximum resultant is 23N , the greater force is

Answer 1

Hey there !!!!!!!

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If F₁ and F₂ are two forces with resultant "R'' and ∅ being angle between them

           R=(√F₁²+F₂²+2F₁F₂cos∅)

If  F₁ and F₂ are at 90°

   R=(√F₁²+F₂²+2F₁F₂cos90)

      =(√F₁²+F₂²)

If R is maximum ⇒⇒cos∅ is maximum

Maximum value of cos∅= 1

So,

 R=(√F₁²+F₂²+2F₁F₂cos0)=(√F₁²+F₂²+2F₁F₂)

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According to question 

Resultant of F₁ and F₂ at right angle is 17N

 17=(√F₁²+F₂²)

  289=(F₁²+F₂²)

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maximum resultant is 23N  

23= (√F₁²+F₂²+2F₁F₂)

 529=(F₁²+F₂²+2F₁F₂)

 289=(F₁²+F₂²)

 529=289+2F₁F₂

 529-289=2F₁F₂

 240=2F₁F₂

 120=F₁F₂

Now 

(F₁+F₂)²=F₁²+F₂²+2F₁F₂

F₁+F₂=√289+240

F₁+F₂=23

Similarly (F₁-F₂)²=F₁²+F₂²-2F₁F₂

F₁-F₂=√289-240

F₁-F₂=√289-240=√49=7

F₁-F₂=7
F₁+F₂=23
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2F₁=30

F₁=15 F₂=8
 
So greater force is 15N

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Hope this helped you............ 

Answer 2

Greater force of P is 15N

Given:

The resultant of 2 forces at right angle is 17N, maximum resultant is 23N

To Find:

The greater force if the maximum resultant is 23 N

Solution:

Given that the resultant of 2 forces at right angles is 17 N, we can write

$P^{2} + Q^2 = 17^2$.............(1)

Again it is also given that the maximum resultant is 23 N.

The resultant of two forces becomes maximum when the angle between them is zero. So the magnitude of the resultant will be sum of the magnitudes of two vectors

So we can also write,

P + Q = 23............(2)

Now,

$(P+Q)^{2} + (P-Q)^{2} = 2(P^2+Q^2)$

$23^{2}$ + $(P-Q)^{2} = 2 (17^{2})$

P - Q = $\sqrt{2( 17^2) -23^2 }$

= 7N

So,

Upon solving,

P -Q =7 ...............(3)

Adding (2) and (3) we get,

2P=30

⇒P=15 N

So greater force P=15N

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