Question

Four forces are acting on a particle. One force of magnitude 3N is directed upward, another is directed 370 East of North having Magnitude 5N, third is directed in South-West direction is of magnitude 4\sqrt 2 N4 2 ​ N and fourth force is \sqrt{5n} N 5n ​ N. If the particle is in equilibrium. Find the value of n.

Answer 1

Answer:

n=2

Explanation:

Refer to the picture below for step by step explanation.

Answer 2

Answer:

The value of n is 2

Explanation:

The forces $F_{2}$ and $F_3$ are on the x-axis whereas the forces $F_1$ and $F_2$ are on the y-axis.

Now, force on the x axis:

$F_x$ = $F_2$ sin$37^o$ + (- $F_3$ sin$45^o$)

$F_x$= 5× 3/5 - (4√2×1/√2)

$F_x$ = -1N

Now, force on the y axis:

$F_y$ = $F_1$ + $F_2$ cos$37^o$ - $F_3$ cos$45^0$

$F_y$ = 3 + 5×4/5 - 4√2×1/√2

$F_y$ = 3N

The net force; =$\sqrt$$(F_x)^2$ + $(F_y)^2$

$F_n_e_t$ = √10

now equating the fourth force to the net force  ( due to the particle are in equilibrium)

$F_4$ = $F_n_e_t$

√5n = √10

n = 2.

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