Question

Three vectors A, B, and C each have a magnitude of 50m and lie in the x-y plane. Their directions relative to the positive x-axis are 30 degrees, 195 degrees, and 315 degrees, respectively. What are (a) the magnitude and (b) the angle of the vector A + B + C, and (c) the magnitude and (d) the angle of A-B+C? What are the (e) magnitude and (f) angle of the fourth vector D such that (A+B)

Answer 1

a)
`R_(A+B+C) =\sqrt{(30.3603)^(2)+(-23.2963)^(2)  } =38.2683 m`

b) Θ=
`tan^(-1) ((R_(y) )/(R_(x) ) )=tan^(-1) ((-23.2963)/(30.3603) )=142.5`º

c)
`R_(A-B+C) =\sqrt{(126.953)^(2)+(2.5856)^(2)  } =126.97 m`

d) β=
`tan^(-1) ((R_(y) )/(R_(x) ) )=tan^(-1) ((2.5856)/(126.953) )=1.16`º

e)
`D=\sqrt{(-5)^(2) +(12.059)^(2) } =13.0545m`

f) ∅=
`tan^(-1) ((R_(y) )/(R_(x) ) )=tan^(-1) ((12.059)/(-5) )=111.66`º

Step-by-step explanation:

First of all, we need to establish our vectors and it's directions:

A=50 m, 30º

B=50 m, 195º

C=50 m, 315º

Now that we have the three vectors, we need to calculate the x and y components:

`A_(x) =50cos(30)=25√(3)m`

`A_(y) =50sin(30)=25m`

`B_(x) =50cos(195)=-48.2963 m`

`B_(y) =50sin(195)=-12.941 m`

`C_(x)=50cos(315)=25√(2)m`

`C_(y)=50sin(315)=-25√(2)m`

Now, that we have the components, we can calculate the resultant's components:

`R_(x) =A_(x) +B_(x) +C_(x)=25√(3)  +(-48.2963)+25√(2) =30.3603m`

`R_(y) =A_(y) +B_(y) +C_(y)=25 +(-12.941)+(-25√(2)) =-23.2963m`

To find the resultant of the vector A+B+C we need to do the following steps:

a)
`R_(A+B+C) =\sqrt{(30.3603)^(2)+(-23.2963)^(2)  } =38.2683 m`

To find the angle it's necessary to use
`tan^(-1)`:

b) Θ=
`tan^(-1) ((R_(y) )/(R_(x) ) )=tan^(-1) ((-23.2963)/(30.3603) )=142.5`º

To find the resultant of the vector A-B+C we need to do the following steps:

`R_(x) =A_(x) -B_(x) +C_(x)=25√(3)  -(-48.2963)+25√(2) =126.953m`

`R_(y) =A_(y) -B_(y) +C_(y)=25 -(-12.941)+(-25√(2)) =2.5856m`

c)
`R_(A-B+C) =\sqrt{(126.953)^(2)+(2.5856)^(2)  } =126.97 m`

To find the angle it's necessary to use
`tan^(-1)`:

d) β=
`tan^(-1) ((R_(y) )/(R_(x) ) )=tan^(-1) ((2.5856)/(126.953) )=1.16`º

To find D=A+B it's important to follow the following steps:

`R_(x) =A_(x) +B_(x) =25√(3)  +(-48.2963)=-5m`

`R_(y) =A_(y) +B_(y) =25 +(-12.941)=12.059m`

e)
`D=\sqrt{(-5)^(2) +(12.059)^(2) } =13.0545m`

f) ∅=
`tan^(-1) ((R_(y) )/(R_(x) ) )=tan^(-1) ((12.059)/(-5) )=111.66`º