Question

Topic : vectors

1] two forces are acting on a single point , if the angle between the two forces (vectors) is 60 degree , and the magnitude of forces are 5N and 10N respectively, find the resultant force.
2] two forces are acting to a body pulling it. if A force is 5N and the B force is also 5N and angle between them is 60 degree find the resultant and the direction of resultant wrt A .

Answer 1

1st Question:

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Given :

• angle between the two vectors = 60°
• magnitude of vector A  = 5N
• magnitude of vector B = 10N

To find :

the magnitude of the resultant force

Formula To Be Applied :

we will here apply the formula of the magnitude of resultant vector :

$\tt{R = \sqrt{A^{2}+B^2+2AB cos\theta }}$

Solution :

given, magnitude of vector A = 5N

magnitude of vector B = 10 N

Theta = 60°

so, resultant force =

$\tt{\implies R = \sqrt{A^2+B^2+2ABcos\theta}}$

$\tt{\implies R = \sqrt{5^2+10^2+2(5)(10)cos60\degree}}$

$\tt{\implies R= \sqrt{25+100+5\times10\times\frac{1}{2}}}$

$\tt{\implies R= \sqrt{125+5\times5}}$

$\tt{\implies R = \sqrt{125+25}}$

$\tt{\implies R = \sqrt{150}}$

$\tt{\implies R = \sqrt{5\times5\times3\times2}}$

$\red{\underline{\boxed{\tt{\therefore \: R = 5\sqrt{6} }}}}$

Hence, magnitude of R = 5√6 N

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2nd question :

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Given :

• Vector A = 5N
• Vector B = 5N
• Angle between them = 60°

To Find :

the magnitude of the resultant and its direction with respect to A

Solution :

given, magnitude of vector A = 5 N

Magnitude of vector B = 5N

angle between them = 60°

$\tt{\implies R = \sqrt{5^{2}+5^2+(5)(5)cos60\degree} }$

$\tt{\implies R = \sqrt{25+25+25\times1/2}$

$\tt{\implies R = \sqrt{52.5}$

hence, the resultant = √52.5 N

As magnitude of vector A and vector B are equal, hence the resultant passes through middle of them

hence, resultant is 30° inclined from A

Answer 2

Answer:

1) resultant will be ✓175

2) resultant is 5 ✓ 3

and angle is 30 degree