Question

33. Five forces act on an object: (1) 60 N at 90°,

(2) 40 N at 0°, (3) 80 N at 270°, (4) 40 N at
180°, and (5) 50 N at 60° What are the magni-
tude and direction of a sixth force that would
produce equilibrium?

Answer 1

See below

Step-by-step explanation:

If you draw the vectors, you will see the 40 N 's cancel out

and you are left with 20 N at 270 and 50 N at 60 (see diagram)

find R... then the 'balancing' R will be of same magnitude in opposite (+180 degrees) direction

x component of 50 N 50 cos 60 = 25 N

y component of 50 N 50 sin 60 = 43.3 N

So now you have 25 N in x direction and (43.3 -20) = 23.3 N at 90 deg

R Magnitude = sqrt( 25^2 + 23.3^2 ) = 34.2 N

R angle = arctan ( 23.3/25) = ~43 degrees

So balancing force would be 34.2 N at (43+180) = 223 degrees