Question

Two forces of 80 N and 70 N act simultaneously at a point. Find the resultant force, if the angle between them is 150°

[Ans. 106.3 N; 61°]

[Source: A textbook of Engineering Mechanics ; R S KHURMI]

Answer 1

Answer :-

R = 40.149 N

Given :-

$F_1 = 80 N$

$F_2 = 70 N$

$\theta = 150^{\circ}$

To find :-

The magnitude of resultant force .

Solution:-

From vector law of Addition we have ,

$\huge \boxed {R = \sqrt{F_1^2 + F_2^2 + 2.F_1.F_2 Cos\theta}}$

Put the given value,

→$R = \sqrt{(80)^2 +(70)^2 + 2\times 80 .70 .Cos150^{\circ}}$

→$R = \sqrt{6400 + 4900 + 11,200 .\dfrac{-\sqrt{3}}{2}}$

→$R = \sqrt{11,300 - 11,200\times \dfrac{\sqrt{3}}{2}}$

→$R = \sqrt{11,300 - 5,600\sqrt{3}}$

→$R = \sqrt{11,300 - 5,600 \times 1.73}$

→$R = \sqrt{11,300 - 9,688}$

→$R = \sqrt{1,612 }$

→$R = 40.149 N$

hence,

The magnitude of resultant is 40.149 N

Answer 2

Answer:

Resultant force = 38.73 N

Explanation:

given that,

Two forces of 80 N and 70 N act simultaneously at a point

given Angie between the two forces = 150°

we know that,

Resultant of two vectors

when angle between the two vectors are given then,

resultant = √(f1² + f2² + 2f1f2cos@)

where,

f1 and f2 are the two forces

and @ is the angle between the two forces

Putting the values,

Resultant force =

√(80² + 70² + 2(80)(70)cos150)

√(6400 + 4900 + 11200 × (-√3/2))

√(11200 - 5600√3)

√(11200 - 9,699.484

= √1,500.516

= 38.73

so,

Resultant force =38.73 N