Question #6ade2

1 Answer
Dec 12, 2017

Magnitude of 7N.
Angle to the x-axis of #51.79^@) to 4sf.

Explanation:

The first thing to do, even if you do it crudely, is to draw a diagram.

enter image source here

I have labelled these vectors ulA and ulB

We can write these vectors in terms of uli and ulj notation, where uli is the horizontal unit vector and ulj is the vertical unit vector. This will make it easier to find the resultant force.

First, lets find ulB since this is easier than ulA.

ulB=-8ulj;
it is not going anywhere horizontally and goes 8 down, or -8 up.

For ulA, we will use trigonometry to find the components.

"vertical"_ulA=5cos30 from SOHCAHTOA.
=5*sqrt3/2
=5/2sqrt3

"horizontal"_ulA=5sin30
=5/2

Its unusual in mechanics that we leave answers as fractions or surds, but it helps us to remain accurate until the end.

:. ulA=5/2sqrt3uli+5/2ulj

Let the resultant force be ulR.

ulR=ulA+ulB
=5/2sqrt3uli+5/2ulj-8j
=5/2sqrt3uli-11/2ulj

enter image source here
Let x be the angle that the resultant force makes with the positive x-axis, or with the unit vector uli.

First, magnitude. By Pythagoras' Theorem:

abs(ulR)=sqrt(((5sqrt3)/2)^2+(11/2)^2
=sqrt(75/4+121/4)
=sqrt(196/4)
=sqrt49
=7

:. absulR=7N

tanx=(11/2)/((5sqrt3)/2)
=11/(5sqrt3)

No point trying to rationalize this, chuck it into your calculator.

x=51.79^@ (4sf)