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JEE Main
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Physics
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Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

There are two forces each having same magnitude 10N. One is...

There are two forces each having same magnitude 10N.10\text{N}\text{.} One is inclined at an angle of 30o{30}^{o} and other is inclined at an angle of 135o{135}^{o} to the positive direction of xx -axis. The xx and yy components of the resultant areThere are two forces each having same magnitude 10N. One is inclined at an angle of 30 ^ o and other is inclined at an

Options

  1. A1.59\text{N}\overset{^}{i} and 12.07\text{N}\overset{^}{j}
  2. B10\text{N}\overset{^}{i} and 10\text{N}\overset{^}{j}
  3. C1 . 59\text{N}\overset{^}{i} and 10\text{N}\overset{^}{j}
  4. D1.59\text{N}\overset{^}{i} and 2\text{N}\overset{^}{j}

Correct answer

A. 1.59\text{N}\overset{^}{i} and 12.07\text{N}\overset{^}{j}

Step-by-step solution

There are two forces each having same magnitude 10N. One is inclined at an angle of 30 ^ o and other is inclined at an
After resolving forces in xy−xy - direction, we get
Along xx direction, net force;
Fx=10[cos30o+(−cos45o)]=10(32−12)=1.59N{F}_{x} = 10\left[cos{30}^{o} + \left(- cos{45}^{o}\right)\right] = 10\left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}}\right) = 1.59\text{N}
Along yy direction, net force;
{F}_{y} = 10 \left(\text\sin{30}^{o} +10\text\sin{45}^{o}\right)<br/>{F}_{y} = 10( 12+12)<br/>\dfrac{1}{2} + \dfrac{1}{\sqrt{2}} )<br/>= 12.07N<br/>Thusforcesin<br/>Thus forces inxandandydirectionaredirection are1.59Nand\text{N} and12.07\text{N}respectively.respectively.
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