Question

A vector of 10 N makes an angle of 45° with x-axis. What will be the angle between its rectangular components?​

Answer 1

Answer:

first of all let's find the 2 forces

F2 = 10sin45

F1 = 10cos45

$f2 = \frac{10}{ \sqrt{2} } = 7.07 \\ f1 = \frac{10}{ \sqrt{2} } = 7.07$

also we can find the resultant force

$r = \sqrt{ \frac{10}{ \sqrt{2} } + \frac{10}{ \sqrt{2} } } \\ r = 10n$

after finding the forces we can find the angle

between them

${r}^{2} = {p}^{2} + {q}^{2} + 2pq \cos( \theta) \\ {10}^{2} = { \frac{10}{ \sqrt{2} } }^{2} + { \frac{10}{ \sqrt{2} } }^{2} + 2 \times \frac{100}{2} \cos( \theta) \\ \cos( \theta) = 0 \\ = 90$

90°