A-plane-is-inclined-at-an-angle-of-30-with-horizontal-Find-the-component-of-a-force-F-10k-N-perpendicular-to-the-plane-Given-that-z-direction-is-vertically-upwards-

Question Number 14923 by Tinkutara last updated on 05/Jun/17

A plane is inclined at an angle of 30 °

with horizontal . Find the component

of a force \vec{F} = - 10 \overset{\land}{k} N perpendicular to

the plane . Given that z - direction is

vertically upwards .

Answered by ajfour last updated on 10/Jun/17

{\bar{F}}_{⊥} = (10 \cos 30 °) (\sin 30 ° \hat{i} - \cos 30 ° \hat{k}) N

= \frac{10 \sqrt{3}}{2} (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{k}) N

= \frac{5 \sqrt{3}}{2} (\hat{i} - \sqrt{3} \hat{k}) N .

∣ {\bar{F}}_{⊥} ∣= 5 \sqrt{3} N .

Commented by Tinkutara last updated on 10/Jun/17

Sir, {wouldn}^{'} t it will be - 5 \sqrt{3} \overset{\land}{k} N ?

Commented by ajfour last updated on 10/Jun/17

q u e s t i o n i s w r o n g l y f r a m e d . .

i f f o r c e i s ⊥ t o i n c l i n e p l a n e, i s

t h e f o r c e a l o n g z a x i s, h o w i s

t h e n t h e i n c l i n e d p l a n e o r i e n t e d ?

Commented by Tinkutara last updated on 10/Jun/17

Why kilo N ? I was saying it would be

- 5 \sqrt{3} N or not ? (minus sign)

Commented by ajfour last updated on 10/Jun/17

i f y o u m e n t i o n t h e m a g n i t u d e

i t i s 5 \sqrt{3} N . (n o m i n u s s i g n)

w h i l e i f t h e c o m p o n e n t o f t h e

v e r t i c a l l y d o w n w a r d f o r c e

p e r p e n d i c u l a r t o t h e i n c l i n e p l a n e

t o b e w r i t t e n i n v e c t o r f o r m, t h e n

{\bar{F}}_{⊥} = 5 \sqrt{3} (\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{k}) .

Commented by Tinkutara last updated on 11/Jun/17

Thanks Sir!