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Answer 1

option d

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Answer 2

Question:- Following sets of forces act on a body. In which case the resultant cannot be zero. ?

(a)10N,10N,10N

(b)10N,10N,20N

(c)10N,20N,23N

(d)10N,20N,40N

Answer:- Option d) 10, 20 ,40

Step by  Step Solution:-

We have  (in a set) 3 forces  acting  on a  body.The resultant can only  be  zero resultant  of any  two forces  will be  equal to third force.

For  following  cases; We will let  that resultant  of  first  two forces is equal to third one and  angle between first  and  second  force is  Ф.

We will solve  for Ф, and  if  we  find an invalid value  of Ф then we  can say that  for that  set  of  forces  resultant cannot  be zero.

If there are two forces P and Q and angles between them is Ф and  their  resultant  is  R,

$R^2=P^2+Q^2+2PQ\cos\theta$

a) For 10 , 10 ,10

$10^{2}=10^2+10^2+2\times10\times10\times\cos\theta\\\;\\0=10^2+200\cos\theta\\\;\\\cos\theta=-\frac{100}{200}\\\;\\\cos\theta=-\frac{1}{2}\\\;\\\theta=\frac{2\pi}{3}$

For  this set we get  a  valid  value of Ф. Therefore, their resultant  can be  zero.

b) 10 , 10 , 20

$20^{2}=10^2+10^2+2\times10\times10\times\cos\theta\\\;\\400=100+100+200\cos\theta\\\;\\400-200=200\cos\theta\\\;\\\cos\theta=\frac{200}{200}\\\;\\\cos\theta=1\\\;\\\theta=0$

For  this set we get  a  valid  value of Ф. Therefore, their resultant  can be  zero.

c) 10 , 20 , 23

$23^{2}=10^2+20^2+2\times10\times20\times\cos\theta\\\;\\529=100+400+400\cos\theta\\\;\\529-500=400\cos\theta\\\;\\\cos\theta=\frac{29}{400}\\\;\\\theta=\cos^{-1}(\frac{29}{400})$

For  this set we get  a  valid  value of Ф. Therefore, their resultant  can be  zero.

d) 10 , 20 , 40

$40^{2}=10^2+20^2+2\times10\times20\times\cos\theta\\\;\\1600=100+400+400\cos\theta\\\;\\1600-500=400\cos\theta\\\;\\\cos\theta=\frac{1100}{400}\\\;\\\theta=\cos^{-1}(\frac{1100}{400})\\\;\\\theta=cos^{-1}(\frac{11}{4})$

For  this set we get  a  invalid  value of Ф. Therefore, their resultant  can not be  zero.

Why here Ф is  invalid? :- We  know that    0 ≤cosФ ≤1 i.e. cosФ can be equal to zero or  greater than zero and less than zero or  equal  to zero. But here we get cosФ=11/4 which is greater than 1