Question

sum of two forces acting at a point is 8 N if the resultant of the forces at right angle to the smaller of two forces have magnitude of of 4N the individual force are ?​

Answer 1

Answe :-

In ∆DAC,

${q}^{2} = {4 }^{2} + {p}^{2}$

Given, P + Q = 8

${q}^{2} - {p}^{2} = {4}^{2}$

(Q - P)(Q + P) = 16

(Q - P)8 = 16

Q - P = 2

Now,

(Q + P) = 8 and (Q-P) = 2

Add, we get 2Q = 10

Q= 5

P = 8-5 = 3

Hence, The forces are 3N and 5N.

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Answer 2

» Sum of two forces acting at a point is 8N.

• Let smaller force be M and greater force be N.

=> M + N = 8

=> M = 8 - N _________ (eq 1)

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» If the resultant of the forces at right angle to the smaller of two forces have magnitude of 4N.

Resultant (R) is perpendicular to the smaller force.

By Pythagoras theoram

=> N² = R² + M²

=> R² = N² - M²

=> (4)² = N² - M²

=> 16 = N² - M²

=> N² - M² = 16

=> N² - (8 - N)² = 16

=> N² - (64 + N² - 16N) = 16

=> N² - 64 - N² + 16N = 16

=> 16N - 64 = 16

=> 16N = 16 + 64

=> 16N = 80

=> N = 80/16

=> N = 5

Put value of N in (eq 1)

=> M = 8 - 5

=> M = 3

$\textbf{Smaller force = 3N and}$

$\textbf{Greater force = 5 N}$

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$\textbf{Individual forces are 3N and 5N}$

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