Question

two boys raising a load pull at an angle to each other. if they exert force of 30N and 60N respectively and their effective is at right angle to the direction of pull of first boy. what is the angle between their arms? what is the effect put.

Answer 1

F1 = 30 N
F2= 60N
resultant force = R

R = √F1² + F2² + 2F1F2 cos0

angle of the resultant force will be the force applied by first boy is 90

tan 90 = B sin 0/ A + B sin 0
infinity = B sin 0/ A + B sin 0
A + B cos 0 = 0
A = - B cosФ
cosФ= -A/ B
= -1/2
Ф= 120

so R = √30² + 60² + 2 x 30 x 60 x cos 120 
R = 51. 96N

Answer 2

The angle between their arms will be 120° and the effective pull will be 51.9 Newtons.

Given: Force applied by two boys 30N and 60N, their resultant is 90° to direction of pull of first boy.

To find: The angle between their arms and the effective pull.

Solution:

Let the force applied by first boy is F1 = 30N.

Let the force applied by first boy is F2 = 60N.

Let the angle between F1 and F2 is A.

Let the resultant of F1 and F2 be R. Then,

Let α be the angle between R and F1.

Given α = 90°

The angle α  between the resultant R and F1 will be given by,

tan α = $\frac{F2 × sin(A) }{F1 + F2 cos(A)}$

tan 90° =  $\frac{ 60 sin(A) }{30 + 60 cos(A)}$

$\frac{1}{0}$ = $\frac{ 60 sin(A) }{30 + 60 cos(A)}$

30 + 60cos(A) = 0

cos(A) = $\frac{-1}{2}$

⇒ A = 120 °

Let the angle between F1 and F2 is 120°.

To find the resultant R,

R = $\sqrt{F1^{2} +F2^{2} + 2F1 F2 cos (A) }$

R = $\sqrt{30^{2} + 60^{2} + 2*30*60*cos(120) }$

R = $\sqrt{900 +3600 -2*30*60*(-1/2)}$

R = 51.96 N

Hence, the angle between their arms will be 120° and the effective pull will be 51.9 N.

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