Mechanics> vectors...

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

nandini , 13 Years ago

Grade 12

7 Answers

Yogita Bang

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.

Approved

Last Activity: 13 Years ago

azeem khan

in 3rd & 4th case.

Last Activity: 13 Years ago

Aravind Bommera

In 4th case

Last Activity: 13 Years ago

hellothisisme

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.

10+10>10
vectors can cancel out each other without other
10+20>30
this is not upheld in option 4 where 10+20= 30 is smaller than the third side i.e. 40

Last Activity: 8 Years ago

Ajeet Tiwari

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

Last Activity: 5 Years ago

Vikas TU

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.

Last Activity: 5 Years ago

Abdul

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