Three forces act on a point: 3 N at 0°, 4 N at 90°, and 5 N at 217°. What is the net force?

1 Answer
Mar 30, 2018

The resultant force is #"1.41 N"# at #315^@#.

Explanation:

The net force #(F_"net")# is the resultant force #(F_"R")#. Each force can be resolved into an #x#-component and a #y#-component.

Find the #x#-component of each force by multiplying the force by the cosine of the angle. Add them to get the resultant #x#-component.

#Sigma(F_"x")=("3 N"*cos0^@) + ("4 N"*cos90^@) + ("5 N"*cos217^@)"="-1 "N"#

Find the #y#-component of each force by multiplying each force by the sine of the angle. Add them to get the resultant #x#-component.

#Sigma(F_y)##=##("3 N"*sin0^@) + ("4 N"*sin90^@) + ("5 N"*sin217^@)"="+1 "N"#

Use the Pythagorean to get the magnitude of the resultant force.

#Sigma(F_R)##=##sqrt((F_x)^2+(F_y)^2)#

#Sigma(F_R)##=##sqrt((-1 "N")^2+(1 "N")^2)#

#Sigma(F_R)##=##sqrt("1 N"^2 + "1 N"^2)#

#Sigma(F_R)##=##sqrt("2 N"^2)#

#Sigma(F_R)##=##"1.41 N"#

To find the direction of the resultant force, use the tangent:

#tantheta=(F_y)/(F_x)=("1 N")/(-"1 N")#

#tan^(-1)(1/(-1))=-45^@#

Subtract #45^@# from #360^@# to get #315^@#.

The resultant force is #"1.41 N"# at #315^@#.