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 F^→ = −10k^∧ N perpendicular to the plane. Given that z-direction is vertically upwards.

Answered by ajfour last updated on 10/Jun/17

F_⊥ ^� =(10cos 30°)(sin 30° i^� −cos 30° k^� )N = ((10(√3))/2)((1/2)i^� −((√3)/2) k^� )N = ((5(√3))/2)(i^� −(√3) k^� )N . ∣F_⊥ ^� ∣=5(√3) N .

Commented byTinkutara last updated on 10/Jun/17

Sir, wouldn′t it will be −5(√3)k^∧ N?

Commented byajfour last updated on 10/Jun/17

question is wrongly framed.. if force is ⊥ to incline plane , is the force along z axis, how is then the inclined plane oriented ?

Commented byTinkutara last updated on 10/Jun/17

Why kilo N? I was saying it would be −5(√3) N or not? (minus sign)

Commented byajfour last updated on 10/Jun/17

if you mention the magnitude it is 5(√3) N.(no minus sign) while if the component of the vertically downward force perpendicular to the incline plane to be written in vector form , then F_⊥ ^� = 5(√3)((1/2)i^� −((√3)/2)k^� ).

Commented byTinkutara last updated on 11/Jun/17

Thanks Sir!