Fundamental Forces Problem: Net Force from Gravity

[JoeyBob]

Homework Statement

See attached

Relevant Equations

F=-Gm_1m_2/r^2

So what I did was find each of the forces the masses had on m1 using the above equation.

From m2 I found 19.975 in the negative i hat and for m4 i found 29.96 in the positive k hat direction using the above equation.

For m3 I used pythagorean theorem to calculate r, which was 3.25 (so r^2 was 10.58). I then used the above equation to get 9.987N. This force would act in the up direction, so I added it to my k hat to get 39.95 and to the left so I made it negative and added it to my i hat to get 29.96 in the negative i hat direction.

Then I squared and added the forces and square rooted them to get 49.937N, which is wrong. The answer is supposed to be 45.84.

I think I screwed up somewhere during the m3 part, but I am not sure how.

 

[Charles Link]

You need to split the force from m3 into components. You just need a little trigonometry for that. To add the whole amount to each of the x and y results from m2 and m4 is incorrect.

 

[JoeyBob]

You need to split the force from m3 into components. You just need a little trigonometry for that. To add the whole amount to each of the x and y results from m2 and m4 is incorrect.

So I divide it by 2 and add it to each component?

 

[Charles Link]

So I divide it by 2 and add it to each component?

That is also not correct. You need to learn about vector addition.
See https://mathworld.wolfram.com/VectorAddition.html

 

[JoeyBob]

That is also not correct. You need to learn about vector addition.
See https://mathworld.wolfram.com/VectorAddition.html

You multiply it by cos(45)?

 

[JoeyBob]

That is also not correct. You need to learn about vector addition.
See https://mathworld.wolfram.com/VectorAddition.html

Wait that doesn't really make sense to me. It gives me the right answer, but the next question is like the same thing but it asks for the angle and the angles isn't 45.

 

[Charles Link]

Splitting the vector into components usually involves taking $\cos{\theta}$ and $\sin{\theta }$.
For your first problem, $\cos (45)=\sin(45)$.

 

[JoeyBob]

Splitting the vector into components usually involves taking $\cos{\theta}$ and $\sin{\theta }$.
For your first problem, $\cos (45)=\sin(45)$.

But how do we know the angle is 45? The second question asks for the angle and its the same type of problem and the angle is 46.79.

 

[Charles Link]

But how do we know the angle is 45? The second question asks for the angle and its the same type of problem and the angle is 46.79.

The force acted along the line joining the m1 and m3.

 

[Delta2]

The force from m3 to m1 acts along the line connecting m3 and m1, that is the diagonal of square , so it makes angle 45 degrees. But this force is not the total force on m1, it is part of the total force. If you have understand that part about vector addition correctly, i think you can understand that the total force (which will be the vector addition of all those three forces) will have another angle.

 

[JoeyBob]

The force from m3 to m1 acts along the line connecting m3 and m1, that is the diagonal of square , so it makes angle 45 degrees. But this force is not the total force on m1, it is part of the total force. If you have understand that part about vector addition correctly, i think you can understand that the total force (which will be the vector addition of all those three forces) will have another angle.

So if I wanted to find the angle for the total force from m1 to m4, I would divide the k hat vector number by the magnitude and take the arccos of this? But this gives me a number that's too high.

 

[Delta2]

Ehm but the total force from m1 to m4, is only that one force right? So you essentially take arccos(1) which is 0, i don't understand what high number you are talking about.

 

[Delta2]

perhaps you mean the force from m2 to m4?

 

[JoeyBob]

perhaps you mean the force from m2 to m4?

Angle the total force on m1 would make from the line from m1 to m4.

 

[Delta2]

Yes well, i see now, you have to divide the total k hat by the magnitude of the total force and then take the arccos of that. The total k hat will be the sum of two components : one from m4 and one from the force from m2. I mean The k hat component of the force from m2.

 

[Delta2]

er sorry i meant m3 instead of m2 in the above.

 

[JoeyBob]

Yes well, i see now, you have to divide the total k hat by the magnitude of the total force and then take the arccos of that. The total k hat will be the sum of two components : one from m4 and one from the force from m2. I mean The k hat component of the force from m2.

Thanks I got it. I suppose it would be faster to just use arctan if you have both vectors, though this wouldn't work with 3 vectors.

 

[Delta2]

You can use arctan as well, you just have to take $$\arctan\frac{\text{total i hat component}}{\text{total k hat component}}$$ to find the angle that makes with the k hat.