Question

The resultant force of 5N nd 10N can not be

1)12N
2)8N
3)4N
4)5N

IF U KNOW THE ANSWER THEN ONLY ANSWER IT OTHERWISE LEAVE IT

Answer 1

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$let \: the \: vectors \: be \: p \: and \: q \\ and \: range \: be \: r \\ so \\ \\ |p - q| \leqslant |r| \leqslant |p + q| \\ using \: this \: concept \\ \: take \: p = 5 \: and \: q = 10 \: so \\ |5 - 10| \leqslant |r| \leqslant |5 + 10| \\ | - 5| \leqslant |r| \leqslant |15| \\ 5 \leqslant |r| \leqslant 15 \\ so \: range \: should \: be \: in \: between \: 5 \: to \: 15$
and in this range 4 is not present.
hence 4 cannot be resultant of the two vectors.

Answer 2

Therefore the resultant force of 5N and 10N cannot be 4N.( Option - 3 )

Given:

The 2 forces are 5 N and 10 N.

To Find:

The option which cannot be a resultant of given 2 forces.

Solution:

This problem can be solved as shown in the process below.

The resultant of the given 2 forces is given by,

⇒ R = √ ( F₁² + F₂² + 2F₁F₂cosθ )

Let F₁ = 5 N and F₂ = 10N

For the resultant to be maximum, θ = 0° ⇒ cosθ = 1

⇒ Rmax = √ ( F₁² + F₂² + 2F₁F₂ )

⇒ Rmax = √ ( 5² + 10² + ( 2 × 5 × 10 ) )

⇒ Rmax = √ ( 25 + 100 + 100 )

⇒ Rmax = √ ( 225 ) = 15

The Maximum Value for the resultant can be 15 N when both the forces are in the same direction.

For the resultant to be minimum, θ = 180° ⇒ cosθ = -1

⇒ Rmin = √ ( F₁² + F₂²  2F₁F₂ )

⇒ Rmin = √ ( 5² + 10² - ( 2 × 5 × 10 ) )

⇒ Rmin = √ ( 25 + 100 - 100 )

⇒ Rmin = √ ( 25 ) = 5

The Minimum value for the resultant can be 5 N when both the forces are in the opposite direction.

All the values of the resultants when the direction between both the forces changes can be between 15N and 5N.

Therefore the resultant force of 5N and 10N cannot be 4N.

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