Question

two mutually perpendicular forces of 8n and 6n acts on the same body of mass 10 kg .find net forces acting on the body .find magnitude of acceleration of body​

Answer 1

Answer:

Here, m=10 kg

The resultant force acting on the body is

F=

(98N)

2

+(6N)

2

=10N

Let the resultant force F makes an angle θ w.r.t. 8N force.

From figure, tanθ=

8N

6N

=

4

3

The resultant acceleration of the body is

a=

m

F

=

10kg

10N

=1ms

−2

The resulatnt acceleration is along the direction of the resulatnt force.

Hence, the resultant acceleration of the body is 1 ms

−2

at an angle of tan

−1

(

4

3

) w.r.t. 8N force.

Explanation:

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Answer 2

Given :-

Mass of the body = 5 kg

Force₁ = 6N

Force₂ = 8N

To Find :-

The direction and the magnitude of the acceleration in it.

Solution :-

We know that,

• m = Mass
• f = Force
• a = Acceleration

The resultant force,

$\underline{\boxed{\sf F=\sqrt{F1^{2}+F2^{2}} }}$

Given that,

Force₁ = 6N

Force₂ = 8N

Substituting their values,

$\sf F=\sqrt{6^{2}+8^{2}}$

$\sf F=10N$

Now, we know

$\underline{\boxed{\sf Acceleration=\dfrac{Force}{Mass} }}$

Given that, Mass = 10 kg and Force = 10N

Substituting their values,

$\sf Acceleration=\dfrac{10}{10}$

$\sf Acceleration=1 \ m/s^{2}$

Thus, acceleration, a = 1 m/s² in the direction of the resultant force.

$\sf tan \ A=\dfrac{6}{8} =0.75$

$\sf A = tan^{-1}(0.75)$

= 37° with 8N Force