Question

Two forces of 4N acts on a body at an angle of 600. find magnitude and direction of the resultant force acting on the body. ​

Answer 1

Answer:

Here si your answer!

Explanation:

Given, two force act on a body at angle 60° , exact force is equal to 4 N

so, magnitude of resultant force is --

\begin{gathered}\sqrt{ {4}^{2} + {4}^{2} + 2 \times 4 \times 4 \cos(60) } \\ = \sqrt{16 + 16 + 32 \times \frac{1}{2} } \\ = \sqrt{32 + 16} \\ = \sqrt{48} = 4 \sqrt{3} newton\end{gathered}

4

2

+4

2

+2×4×4cos(60)

=

16+16+32×

2

1

=

32+16

=

48

=4

3

newton

now, let angle between resultant force and vector force is ∅

now,

tan ∅ =

\begin{gathered}\frac{4 \sin(60) }{4 + 4 \cos(60) } = \frac{2 \sqrt{3} }{6} = \frac{ \sqrt{3} }{3} \\ = \frac{1}{ \sqrt{3} }\end{gathered}

4+4cos(60)

4sin(60)

=

6

2

3

=

3

3

=

3

1

so, ∅ = 30°

hence, resultant force is 30° from forces 4N

hope it help you

Answer 2

Answer:

above answer is perfectly correct

Explanation:

wish you a cushy life path strewn with roses buddy ♡