Question

Chapter 01, Problem 66 

Three forces are applied to an object, as shown in the figure. Force F1 has a magnitude of 21.8 newtons (21.8 N) and is directed 30.0° to the left of the ty axis. Force F2 has a magnitude of 14.3 N and points along the +x axis. What must be the (a) magnitude and (b) direction (specified by the angle 6 in the drawing) of the third force F3 such that the vector sum of the three forces is 0 N? 

(a) Number _______ Units _______ 

(b) Number _______ Units _______ 

Answer 1

Answer 2

Lets resolve the vector F1 into components

angle from positive x axis, θ = 120

F1x = F1 * cos θ

= 21.8 * cos(120.0)

= -10.9 N

F1y = F1 * sin θ

= 21.8 * sin(120.0)

= 18.8794 N

we have:

F1 = -10.9i+ 18.8794j

F2 = 14.3i

R = F1 + F2

R = (-10.9i+ 18.8794j) + (14.3i)

= (F1x + F2x)i + (F1y + F2y)j

= (-10.9 + 14.3)i + (18.8794 )j

= 3.4i+ 18.8794j

We need to find F3 such that,

F1 + F2 + F3 = 0

So,

R + F3 = 0

F3 = -R

= - (3.4i+ 18.8794j)

= -3.4 i - 18.8794j

magnitude of F3 = sqrt (-3.4^2 + -18.8794^2)

magnitude of F3 = 19.1831 N

Since x component is negative and y component is also negative

It lies in the 3rd cordinate

angle = 180 + atan(|y/x|)

angle = 180 + atan(|-18.8794/-3.4|)

angle = 190.2 degree

so,

thetha = 190.2 - 180 = 10.2 degree

Answer:

19.18 N

10.2 degree