Question

Two force vectors of same magnitude are arranged in the
following manners :
1)90
2)120°
3)60°
Magnitude of resultant force is maximum for:
(a) 1
(c) 2
(d) 2 and 3
(b) 2
pls give full solution.... urgent​

Answer 1

Answer:

d

Explanation:

let magnitude be F

When Both are at 90,

Resultant =F× root 2= 1.414F

When at 120 R= F

When at 60 R=F× root 3=1.732F

Answer 2

The magnitude of the resultant force is maximum for (d) 3.

Given:

Two vectors of magnitude F and F.

To find:

Magnitude of the resultant force at the given angles.

Solution:

For given any two vectors, A and B, the magnitude of the resultant force of the two angles is given by,

$|F| = \sqrt{(A)^{2}+(B)^{2}+2ABcos\alpha }$

Where, $\alpha$ is the angle between the two forces A and B.

Hence,

According to the question, we have been given that the two forces are equal . Lets say F.

Step 1

For $\alpha = 90^{o}$

$|F|=\sqrt{(F)^{2}+(F)^{2}+2(F)(F)cos(90) }$

     $=\sqrt{F^{2}+ F^{2}+ 2F^{2}(0) }$

     $=\sqrt{2F^{2} } = \sqrt{2} F$

Step 2

For $\alpha =120^{o}$

$|F|=\sqrt{F^{2} +F^{2}+2F^{2}cos(120) }$

     $=\sqrt{2F^{2}(1+cos(120^{o} )) }$

     $=\sqrt{2F^{2}(2cos^{2}(60^{o} ) )}$

     $=\sqrt{2F^{2} (2)(\frac{1}{2} )^{2} } =F$

Step 3

For $\alpha = 60$

$|F|=\sqrt{F^{2}+ F^{2}+2(F)(F)cos(60^{o} ) }$

    $=\sqrt{F^{2}+ F^{2}+ 2F^{2}(\frac{1}{2} ) }$

     $=\sqrt{3F^{2} } =\sqrt{3}F$

Final answer:

Hence, the magnitude of forces for angles 90°, 120°, 60° are $\sqrt{2} F$ ,F and $\sqrt{3}F$ respectively.

Therefore, the magnitude is the greatest for (d) 3 that is equal to $\sqrt{3} F$.

Although your question is incomplete, you might be referring to the question below.

Two force vectors of same magnitude are arranged in the following manners.

1. 90°
2. 120°
3. 60°

Magnitude of resultant force is maximum for:

(a) 1

(b) 2

(c) 2 and 3

(d) 3