Question

. The following forces act at a point :
1) 20 N inclined at 30° towards North of East,
i) 25 N towards North,
i) 30 N towards North West, and
) 35 N inclined at 40° towards South of West.
Find the magnitude and direction of the resultant force.
Solution. The system of given forces is shown in Fig​

Answer 1

Answer:

1803.80 N

Explanation:

I hope this will help you ✌️

Answer 2

Given: The following forces act at a point :

1) 20 N inclined at 30° towards North of East,

i) 25 N towards North,

i) 30 N towards North West, and

) 35 N inclined at 40° towards South of West.

To find: the magnitude and direction of the resultant force

Solution: for finding the resultant of the given forces we need to dind resultant of forces in each direction.

Doing the components of the forces in x-direction we get forces as

20cos30°, 25cos90°, 30cos45°, 35cos40°

net force in the x-direction can be written as Fx

Fx = 20cos30°+25cos90°- 30cos45°- 35cos40°

Fx = - 30.7 N

Doing the components of the forces in y-direction we get forces as

20sin30°, 25sin90°, 30sin45°, 35sin40°

Net force in the y- direction can be written as Fy

Fy = 20sin30°+ 25sin90°+30sin45- 35sin40°

Fy = 33.7N

Resultant force will be

F = √( Fx^2 + Fy^2)

F = 45.6N

Direction of resultant force will be

a = tan^-1( 33.7/-3.7)

a = -47.7°

Therefore, the magnitude and direction of the resultant force is 45.6N and -47.7° respectively.