How Do You Find the Gravitational Field at a Corner of a Square?

[Stuffy01]

Simple Gravitation - With Answer - Help me work the middle steps

Homework Statement

Three objects -- two of mass m and one of mass M -- are located at three corners of a square of edge length l as in Figure P13.23. Find the gravitational field g at the fourth corner due to these objects. (Express your answers in terms of the edge length l, the masses m and M, and the gravitational constant G).

Homework Equations

Find the magnitude and direction of the gravitational field g.

The Attempt at a Solution

The direction is 45 degrees (measured counterclockwise from the positive x-axis)

G((4m+M)/(2l^2)) was the final simplified answer I got for the magnitude, but it was wrong.

 

[Stuffy01]

Associated work:

Base equation: Gravitational field force = GMm/(l^2)

I see three gravitational forces at work:
Gm/l^2 on the y-axis

Gm/l^2 on the x axis

GM/((sqrt(2) times l)^2)

Add all three so 2Gm/2l^2 + 2Gm/2l^2 + GM/2l^2 = (G(4m+M))/(2((l)^2))

 

[denverdoc]

Welcome to PF.

Careful, as in Why do all the denominators look alike when you showed the diagonal object to have longer radius?

Never mind i see what you did.

 

[Stuffy01]

Thanks. See any easy mistakes or did I go the wrong direction?

 

[denverdoc]

No I found no mistakes so I was sort of head scratching on this one, hoping someone else might see a problem if there is one. Of course these are vector fields, which might be the issue.

 

[Stuffy01]

I GOT THE ANSWER FROM A FRIEND:

(G/l^2)(M/2+sqrt(2)m)

Can anyone help me work it out now that the answer is known?

 

[Flux = Rad]

It is to do with the fact that the fields are vector fields.

You can't just arithmetically add the 3 fields up, you have to consider their direction. It's like if you push a block north with 1N and east with 1N the overall force is $$\sqrt{2}N$$ in the NE direction, not 2N. So I'd recommend drawing a diagram and doing a bit of pythagoras.

 

[denverdoc]

No problem, resolve the diagonal force into x and y components which can be added like scalars

cos and sin are both sqrt(2)/2 so mult your diag. expression and add to the x and y components from other forces, then calculate magnitude by taking the sqrt of sum of squares. Falls out like honey in a bucket. I suspect you'll never make this mistake again. good problem and post.

 

[Saudi Arabia]

Hi Guys, i guess this is the answer, I'm not sure! but? if someone has a comment reply pls.xxx
tanθ = gy / gx
= [ Gm / L2 + G M /2L2 sinθ ] / [ Gm /L2 + G M / 2L2 cosθ ]

θ = tan-1 { [ Gm / L2 + G M / 2L2 sinθ ] / [ Gm /L2 + G M / 2L2 cosθ ]}