Question

Two forces of magnitude equal to 2p and p respectively act on a particle.if the first is doubled and second is increased by 12N,the direction of resultant is unaltered then value of P

Answer 1

Given:

f1 = 2p

f2 = p

The direction of resultant not changed

To find:

p

Solution:

the direction of resultant given by

tanβ = $\frac{f2sin\alpha }{f1+f2cos\alpha }$

tanβ = $\frac{p sin\alpha }{2p+pcos\alpha }$......(i)

Now, f1 is doubled and f2 increased by 12.

f1 = 4p

f2 = p + 12

now new direction given by

tanβ = $\frac{(p+12) sin\alpha }{4p+(p+12)cos\alpha }$ .....(ii)

as direction is unaltered,

equation (i) and (ii) are equal

$\frac{p sin\alpha }{2p+pcos\alpha }$ = $\frac{(p+12) sin\alpha }{4p+(p+12)cos\alpha }$

4p²sinα + p²sinα cosα + 12psinα cosα = 2p²sinα + 24p sinα + p²sinα cosα + 12psinα cosα

2p²sinα = 24p sinα

2p = 24

p = 12

Answer:

p = 12N

Answer 2

The value of p is 12.

Given,

Two forces of magnitude equal to 2p and p respectively act on a particle. The first is doubled and the second is increased by 12N.

To Find,

The value of p.

Solution,

The direction of resultant given by

tanβ = (f2sinα)/(f1+f2cosα)

tanβ = psinα/(2p+pcosα)----(i)

Now, f1 is doubled and f2 increased by 12, so

f1 = 4p

f2 = p + 12

Now new direction will be

tanβ =  (p+12)sinα/(4p+(p+12)cosα)----(ii)

as direction is unaltered,

equation (i) and (ii) are equal

So,

4p²sinα + p²sinα cosα + 12psinα cosα = 2p²sinα + 24p sinα + p²sinα cosα + 12psinα cosα

2p²sinα = 24p sinα

2p = 24

p = 12

Hence, the value of p is 12.

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