Question

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions parallel to and normal to the plane that makes an angle of theta equals tangent Superscript negative 1 Baseline (StartFraction StartRoot 3 EndRoot Over 3 EndFraction )θ=tan−1 3 3 with the positive​ x-axis. Show that the total force is the sum of the two component forces. What is the component of the force parallel to the​ plane? left angle nothing comma nothing right angle , What is the component of the force perpendicular to the​ plane? left angle nothing comma nothing right angle , Find the sum of these two forces. left angle nothing comma nothing right angle

Answer 1

Let
`$v_0$` be the unit vector in the direction parallel to the plane and let
`$F_1$` be the component of F in the direction of
`v_0` and
`F_2` be the component normal to
`v_0`.

Since,
`|v_0| = 1,`

`$(v_0)_x=\cos 60^\circ= (1)/(2)$`

`$(v_0)_y=\sin 60^\circ= (\sqrt 3)/(2)$`

Therefore,
`v_0 = \left<(1)/(2),(\sqrt 3)/(2)\right>`

From figure,

`|F_1|= |F| \cos 30^\circ = 10 * (\sqrt 3)/(2) = 5 \sqrt3`

We know that the direction of
`F_1` is opposite of the direction of
`v_0`, so we have

`$F_1 = -5\sqrt3 v_0$`

`$=-5\sqrt3 \left<(1)/(2),(\sqrt3)/(2) \right>$`

`$= \left<-(5 \sqrt3)/(2),-(15)/(2) \right>$`

The unit vector in the direction normal to the plane,
`v_1` has components :

`$(v_1)_x= \cos 30^\circ = (\sqrt3)/(2)$`

`$(v_1)_y= -\sin 30^\circ =- (1)/(2)$`

Therefore,
`$v_1=\left< (\sqrt3)/(2), -(1)/(2) \right>$`

From figure,

`|F_2 | = |F| \sin 30^\circ = 10 * (1)/(2) = 5`

∴
`F_2 = 5v_1 = 5 \left< (\sqrt3)/(2), - (1)/(2) \right>`

`$=\left<(5 \sqrt3)/(2),-(5)/(2) \right>$`

Therefore,

`$F_1+F_2 = \left< -(5\sqrt3)/(2), -(15)/(2) \right> + \left< (5 \sqrt3)/(2), -(5)/(2) \right>$`

`$=<0,- 10> = F$`