Net force of two forces in a ring

In summary, the problem involves finding the forces F1 and F2 in a ring with a given angle of 30°, in order for the net force to be 10^3 N downwards without any horizontal component. Two solutions are presented, with different equations for finding F2, resulting in different values. The error in one solution is attributed to a mistake in the equation.
  • #1
ShizukaSm
85
0

Homework Statement


Consider the following ring:
DELETEME.JPG

Considering that [itex]\theta = 30°[/itex], determine F1 and F2 in order for the net force to be oriented downards with magnitude 10^3 N and without any horizontal component.

The Attempt at a Solution


Alright, so, the problem is I find two conflicting solutions when solving this simple problem.

Solution 1:
[tex]
\\
\sum F_x = 0 \rightarrow F_1*Sin(20)-F_2*Sin(30)=0 \\
\sum F_y = 0 \rightarrow F_1*Cos(20)+F_2*Cos(30)=1000
\\ \rightarrow F_2 ({\frac{Sin(30)}{Sin(20)} + Cos(30)}) = 1000
\\ \rightarrow F_2 = 429.56N | F_1 = 627.91N
[/tex]
PS: How do I fix my parenthesis in line 3 of my latex equation? It's lacking proportion.

Solution 2(Book solution):
Solution2.JPG


So... In short, why aren't those two solutions agreeing?
 
Last edited:
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  • #2
ShizukaSm said:
[tex]
\\
\sum F_x = 0 \rightarrow F_1*Sin(20)-F_2*Sin(30)=0 \\
\sum F_y = 0 \rightarrow F_1*Cos(20)+F_2*Cos(30)=1000
\\ \rightarrow F_2 ({\frac{Sin(30)}{Cos(20)} + Cos(30)}) = 1000
[/tex]
Check that last step.
PS: How do I fix my parenthesis in line 3 of my latex equation? It's lacking proportion.
Try using \left( etc. instead of just (.
 
  • #3
haruspex said:
Check that last step.

Try using \left( etc. instead of just (.

Ahh, got it, can't believe I made such a stupid mistake. I checked and re-checked many times but couldn't spot it myself.

If anyone's wondering, it was supposed to be [itex]\frac{F_1*(Sin(30)*Cos(20)}{Cos(20)}+...[/itex]
 
Last edited:

1. What is the net force of two forces in a ring?

The net force of two forces in a ring is the overall force acting on the object in the direction of the resultant force. It is the sum of all the forces acting on the object.

2. How do you calculate the net force of two forces in a ring?

To calculate the net force, you need to determine the magnitude and direction of each force. Then, use vector addition to find the resultant force. The magnitude of the resultant force is the sum of the magnitudes of the individual forces, and the direction is the angle between the two forces.

3. What happens if the two forces are in the same direction?

If the two forces are in the same direction, the resultant force will be the sum of the two forces, and the direction will remain the same. This will result in a stronger net force on the object.

4. Can the net force of two forces in a ring ever be zero?

Yes, the net force of two forces in a ring can be zero if the two forces are equal in magnitude but opposite in direction. This is known as a balanced force, and it means that the object will remain at rest or continue moving at a constant velocity.

5. How does the distance between the two forces affect the net force in a ring?

The distance between the two forces does not affect the net force in a ring. The only factors that affect the net force are the magnitude and direction of the forces. However, the distance between the forces can affect the torque (rotational force) on the object.

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