# Solved Example

SOLED EXAMPLE

Example 1.1. Find the resultant of F1 and F2 forces.

Solution :

R2 =F12 +F22 +2F1F2 cos 300

=(12)2+(9)2+2 x 12 x 9 cos 300

=412.0614

R =√412 .0614= 20.3 N

θ=tan-1 (f2 sin 300/f1+f2 cos 300)= tan -1 (9 sin 300/12+9 cos 300) =12.810

Example 1.2. If the resultant of two equal forces of, magnitude of each F is 30J3 N, find magnitude of F if the angle between F and F is 60°.

Solution.

R2 =F2+F2=2F x F cos 600

(30√3)2  = 2F2 (1+cos 600)

900 x 3/3 =f2 or f =30N .

Example 1.3. Find F1 and f2for Fig. P-1.3 (a) use Lami’s theorem

Solution.

Using Lami’s theorem, we get

F1 /sin 1500 =f2 /sin 1500 =20 /sin 600

F1 =20 sin 1500 /sin600 =11.55 =f2

F1 =f2 =11.55 N

Example 1.4. Find the magnitude of resultant of four forces F1, F2, F3 and F4 acting in order of sides of square. (Ref Fig P-1.4 (a)

R1 =√(20)2 +(30)2 = 36 .05

Tan θ 1 =30/20 ,θ1 =56.310

R2 =√(40)2 +(50)2 =64.03

Θ2 =tan -1 50/40 =51.340

α= 51.34 + 90° + 33.69°

= 175.03°.

R2 = R12 + R22 + 2R1R2 cos 175.03°

= (36.05)2 + (64.03)2 + 2 x  36.05 x 64.03 cos 175.03°

R = √ 800 = 20 √2 N Ans.

Alternate Method

R =√(20)2 +(20)2 =√800 =20 √2 N

Tan θ1 =-20/-20 =1 θ1 =450

. . Direction of R with x axis

=1800 +450 =2250