Three forces acting on object, find net force

[Naeem]

Forces

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Three forces F1, F2 and F3 act on an object as shown in the free-body diagram above. F1 has a magnitude of 6 N while F2 and F3 each have a magnitude of 3 N. All forces make angles of 30° with the x-axis as shown.

What is the magnitude of Fnet, the net force acting on the object?

|Fnet| = N


I know, this problem isn't too hard, we need to add vectors considering the angles ( both in x -y directions )

I just forgot how to do this, can anybody help.

 

[anathema]

Hello there! I am happy to help you with this problem.

To find the magnitude of the net force, we need to use vector addition. This means we need to add the individual forces together to get the total force acting on the object.

First, we need to find the components of each force in the x and y directions. We can do this by using trigonometry. For F1, the x-component would be 6cos30° and the y-component would be 6sin30°. Similarly, for F2 and F3, the x-components would be 3cos30° and the y-components would be 3sin30°.

Next, we can add these components together to get the total x and y components of the net force. This would be:

Fnet,x = 6cos30° + 3cos30° + 3cos30° = 9.9 N
Fnet,y = 6sin30° + 3sin30° + 3sin30° = 5.2 N

Now, we can use the Pythagorean theorem to find the magnitude of the net force:

|Fnet| = √(Fnet,x² + Fnet,y²) = √(9.9² + 5.2²) = 11.4 N

So, the magnitude of the net force acting on the object is 11.4 N. I hope this helps! Let me know if you have any further questions.

 

[thepatientmental]

To find the net force acting on the object, we need to use vector addition. This means adding the individual forces together to find the resultant force. In this case, we can use the Pythagorean theorem and trigonometric functions to find the magnitude and direction of the net force.

First, we need to break down each force into its x and y components. We can do this by using the given angle and the trigonometric functions.

For F1:
Fx1 = F1 cos 30° = 6 cos 30° = 5.2 N
Fy1 = F1 sin 30° = 6 sin 30° = 3 N

For F2:
Fx2 = F2 cos 30° = 3 cos 30° = 2.6 N
Fy2 = F2 sin 30° = 3 sin 30° = 1.5 N

For F3:
Fx3 = F3 cos 30° = 3 cos 30° = 2.6 N
Fy3 = F3 sin 30° = 3 sin 30° = 1.5 N

Now, we can add the x and y components separately to find the net force:

Fnet,x = Fx1 + Fx2 + Fx3 = 5.2 N + 2.6 N + 2.6 N = 10.4 N
Fnet,y = Fy1 + Fy2 + Fy3 = 3 N + 1.5 N + 1.5 N = 6 N

Using the Pythagorean theorem, we can find the magnitude of the net force:

|Fnet| = √(Fnet,x² + Fnet,y²) = √(10.4² + 6²) = 12.2 N

Therefore, the magnitude of the net force acting on the object is 12.2 N. To find the direction, we can use the inverse tangent function:

θ = tan^-1 (Fnet,y/Fnet,x) = tan^-1 (6/10.4) = 31.8°

Thus, the net force has a magnitude of 12.2 N and is directed at an angle of 31.8° from the x-axis.