ThreePeople,LabeledA,B,&C,ArePullingOnASledAsShown.CalculateTheNetforceOnTheSled.

Question

                B \

                     \   20 N

                       \       

                         \     __________

                           \  |                  |   30N

                 30 deg\  |                  |______\   

          ———————|                  |           /  A             THIS IS THE PICTURE. ALSO CALCULATE THE RESULTANT NET

                 60 deg / |_________ |                                   FORCE AFTER FINDING THE COMPONENT NET FORCE IN THE 

                          /                                                         X-DIRECTION AND IN THE Y-DIRECTION.

                        /  50 N

                      /

                    C

Answer 1

You need to break up each force by their x and y components.

 

For force B:

 

F_Bx = 20cos(180 - 30)

F_By = 20sin(180 - 30)

 

 

For force C:

 

F_Cx = 50cos(180 + 60)

F_Cy = 50sin(180 + 60)

 

 

Force A is a horizontal force going in the positive x-direction.

 

F_Ax = A    ---->  Your question did not specify what A was, so I leave it as this.

F_Ay = 0

 

 

Next, combine all the x components and all the y components to get the components net force.

 

R_x = F_Ax + F_Bx + F_Cx

 

R_y = F_Ay + F_By + F_Cy

 

 

 

The resultant vector is

 

R = √(R²_x + R²_y)

 

 

 

Answer 2

Still the force A is missing.  If it is unknown then the object is not in equilibrium! Please give more information. Thanks

Now that you completed the problem my answer will be:

 

you have 3 forces 20 N , 50 N and 30 N.

Drawing the free body diagram in the x direction and y direction you get:

Σ F_x = 30 -20 cos 30 - 50 cos 60

= 30 - 17.32 -25

= -12.32 N this is the resultant in the negative x axis.

and

Σ F_y = 20 sin 30 - 50 sin 60 = 10-43.3

= -33.3 N

The resultant is a vector ‹ -12.32, -33.3 ›

 

Its magnitude is:

((-12.32)² + (-33.3)²)^1/2 =  35.5 N

The direction is given by the angle θ = arc tan (-33.3/-12.32) = 69.7°  in the III quadrant or θ = 69.7 + 180 = 249.7°