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following sets of the three forces act on a body,. In which case resultant cannot be zero? 1) 10N, 10N, 10N 2)10N, 10N, 20N 3)10N, 20N, 20N 4010N, 20N, 40N

following sets of the three forces act on a body,. In which case resultant cannot be zero?


1) 10N, 10N, 10N


2)10N, 10N, 20N


3)10N, 20N, 20N


4010N, 20N, 40N

Grade:12

7 Answers

Yogita Bang
39 Points
11 years ago

Dear Nandini,

In 1st case, if 10N and 10N are at 1200 then resultant of these two is 10N. Then the 3rd force if acted opposite to the resultant then net force is zero.

In 2nd case, if 10N and 10N are in same direction and 20N is opposite to both of them then resultant is zero.

In 3rd case, if 10N and 20N are acted at an angle of cos-1(-1/4) then the resultant is 20N. Then the third force if acted opposite to the resultant then net force is zero.

Explanation of third case: R2 = P2+Q2+2PQcosΘ ; (20)2 = (10)2+(20)2+2*10*20*cosΘ

Therefore cosΘ = (400 - 500)/400 = -1/4. This is possible value of cosΘ as it lies in the interval [-1,1].

Any such situation is not possible in 4th case. Hence, the resultant cannot be zero in 4th case.

 

azeem khan
32 Points
11 years ago

in 3rd & 4th case.

Aravind Bommera
36 Points
11 years ago

In 4th case

hellothisisme
11 Points
6 years ago
for the vectors to have a zero resultant, they must form a complete polygon. in this case a complete triangle must be formed and for a triangle to be formed, sum of any 2 sides is greater than the third one. 
  1. 10+10>10
  2. vectors can cancel out each other without other
  3. 10+20>30
  4. this is not upheld in option 4 where 10+20= 30 is smaller than the third side i.e. 40
Ajeet Tiwari
askIITians Faculty 86 Points
3 years ago
Hello student

In 1st case, if 10N and 10N are at 1200then resultant of these two is 10N. Then the 3rd force if acted opposite to the resultant then net force is zero.

In 2nd case, if 10N and 10N are in same direction and 20N is opposite to both of them then resultant is zero.

In 3rd case, if 10N and 20N are acted at an angle of cos-1(-1/4) then the resultant is 20N. Then the third force if acted opposite to the resultant then net force is zero.

Explanation of third case: R2= P2+Q2+2PQcosΘ ; (20)2= (10)2+(20)2+2*10*20*cosΘ

Therefore cosΘ = (400 - 500)/400 = -1/4. This is possible value of cosΘ as it lies in the interval [-1,1].

Any such situation is not possible in 4th case. Hence, the resultant cannot be zero in 4th case.

Hope it helps
Thankyou and Regards
Vikas TU
14149 Points
3 years ago
let us considered three forces in equilibrium and represents three sides of a triangle a,b and c. We know that, In the triangle, addition of the two sides is greater than the third side then the resultant might be zero but when addition of the two sides is less than the third side then the resultant can not zero.
Abdul
13 Points
2 years ago
In case 4 (to do an do if so ye so it did do if so if so do it did uchit ye uf uf 4th it TV but your try be ch hy fight to if so if the log so hai vo so so if so ch HD so so if we kh co so if so no is ek of eg ch TV ek on TV TV ex ed Dec TV ok on in uf ta ex QC eg UK o on if uh TV ek kk)
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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