Two forces act at an angle of 120 degrees. The bigger force is 40 N and the resultant is perpendicular to the smaller one. What is the smaller force?

 Two forces act at an angle of 120 degrees. The bigger force is 40 N and the resultant is perpendicular to the smaller one. What is the smaller force?

Solution

start with the smaller unknown force of BD

ADB = 120 degrees & AD is an = force of BD

BD has a virtual force of BC + CD, C being the midpoint of AB

AD also has a virtual force of AC + CD

AC cancels BC & CD is doubled to CE, hence DE = CD

DF is perpendicular to BD and the bigger force HD that is along the AD line + the smaller BD will end up somewhere on the DF line.

since BD and AD will end up at E, the remainder of the bigger force will venture off of E and will be parallel to AD & HD as no other forces are apposing it. F is where this parallel line crosses the perpendicular DF line. EF will = HA.

ADB =120, AD=BD and C is the midpoint of A&B, both ADC & BDC are 60 degrees each.

because of the 30/60/90 triangle, CD is 1/2 of AD & BD

also ADE & BDE are = & 120 degrees each (360–120)/2

since BDE = 120 & BDF being perpendicular at 90, EDF = 30 degrees

because AD & EF are parallel, and ADE = 120 degrees, DEF also = 120 degrees

triangle DEF has 2 angles 30 & 120 leaving 30 remaining for DFE (180-(120+30))

EDF and DFE both = 30 degrees so triangle DEF is isosceles & thus DE = EF

CD = DE = EF = HA means EF & HA are also 1/2 of AD & BD

smaller force AD + 1/2 AD (HA) = 40

smaller force = 40 / (1+.5) 

or 40/1.5 or 40/(3/2)

 or 40*(2/3) or 80/3 or 26 2/3