What Fraction Of Her True Weight Was Indicated?

[Stanc]

Homework Statement

Julie tightened herself fastened a set of bathroom spring scales to a platform and went down a 30 degree slope. No friction was present and the platform supporting the scales was HORIZONTAL. what fraction of her true weight was indicated by the scales?

The Attempt at a Solution

so what i did was just assume a random mass, in this case 50kg so that means around 500 Newtons.
Therefore, Fnormal would be 500 cosine 30 = 433 Newtons
than all i did was take 433 and divide it by 500 which gives me 0.866. So is it right to say that 87% of her true weight will be shown on the scale?

 

[PhanthomJay]

Homework Statement

Julie tightened herself fastened a set of bathroom spring scales to a platform and went down a 30 degree slope. No friction was present and the platform supporting the scales was HORIZONTAL. what fraction of her true weight was indicated by the scales?

The Attempt at a Solution

so what i did was just assume a random mass, in this case 50kg so that means around 500 Newtons.
Therefore, Fnormal would be 500 cosine 30 = 433 Newtons
than all i did was take 433 and divide it by 500 which gives me 0.866. So is it right to say that 87% of her true weight will be shown on the scale?

That is not correct. The normal force you have calculated is the force of the plane on the wheels , acting perpendicular to the incline. And that would be the scale force indication if the scale was also parallel to the inclined slope. But it is given that the scale platform is horizontal. You've got to look at the forces acting on the person in the vertical and horizontal directions. Is there acceleration in the vertical direction?

 

[Stanc]

Well there is no acceleration given but can't you say that 9.81m/s^2 is the acceleration due to gravity? and if you do the forces acting in the vertical direction wouldn't it just be 500N up and than fg = 500 N down?

 

[PhanthomJay]

Well there is no acceleration given but can't you say that 9.81m/s^2 is the acceleration due to gravity?

well yes, but the acceleration due to gravity is not the vertical acceleration of the person and scale and platform

and if you do the forces acting in the vertical direction wouldn't it just be 500N up and than fg = 500 N down?

No, that would be true only if the net force in the vertical direction was 0 (no acceleration in the vertical direction). But there is acceleration in the vertical direction, because the person is moving vertically downward at increasing velocity due to a net force downward in that vertical direction. The person is also accelerating horizontally, because he/she is also moving horizontally at increasing speed. Note further that the person's resultant acceleration is parallel to the slope, so you might want to find the acceleration of the platform and person and scale down the plane in the usual manner you are accustomed to for forces acting parallel to the plane. Then find the vertical component of that acceleration and use Newton 2 in the vertical direction.

 

[Stanc]

So the acceleration parallel to the surface is just mg sin theta correct? So, if assuming a random mass of 10kg than acceleration parallel to the surface = 10(9.81) sin 30 = 49.05m/s^2 - isn't that too fast? or should i not assume a weight at all? And how would i find acceleration vertically?, is it (assuming mass of 10kg) mgcos30?

 

[PhanthomJay]

So the acceleration parallel to the surface is just mg sin theta correct?

The force parallel to the incline is mg sin theta. The acceleration parallel to the incline is ___?___

So, if assuming a random mass of 10kg than acceleration parallel to the surface = 10(9.81) sin 30 = 49.05m/s^2 - isn't that too fast?

correct this once you determine the correct value for the acceleration down the incline (call it a_p)

or should i not assume a weight at all?

It's OK to assume it, or just calling it 'm' is better

And how would i find acceleration vertically?, is it (assuming mass of 10kg) mgcos30?

No, the vertical component of the acceleration, a_y, comes from the trig... a_y/a_p = ___?___ (it is not the cos function, draw a sketch)...

 

[Stanc]

Oh you my bad mg sin theta = the force so in order to find acceleration it would be
mg sin theta/m. So you can actually just cancel out mass i guess right? so a = 49.05/10 = 4.905m/s^2.
so...

ap = 4.905m/s^2
is ay than just Fnet/m? Or is it just 9.81m/s^2..im still not too sure about the vertical acceleration...

 

[Stanc]

Wait i think i just got it. With the angle 30 degrees the ay = sin30 = ay/ap = sin30 = ay/4.905m/s^2

that solves to be 2.4525m/s^2 = ay

I can also solve for ax which would equal cos30 = ax/ap = ax /4.905 = 4.24m/s^2

So acceleration in the horizontal equals 4.24m/s^2 and the vertical acceleration = 2.4525m/s^2that is parallel to the surface = 4.905m/s^2

I am honestly clueless however to find the fraction of her true weight

 

[Stanc]

Wait to solve for the fraction of the true weight, would i take acceleration in the y (2.4525m/s^2) and subtract it from gravity so 9.81 - 2.4525 = 7.3575m/s^2 and than multiply that by 10 which gives me 72.2 Newtons. And than would i take 9.81 x 9.81 = 96.2361 Newtons and than divide 72.2/96.2361 = roughly 3/4 her true weight?

 

[Stanc]

Wait would i do the 9.81 x 9.81 part??

 

[azizlwl]

I think you have to compare this with a girl inside an elevator.
If accelerating upward the weight increases, N-mg=a
In your case find downward acceleration and you will get N, the weight.

 

[Stanc]

Ya I found downward acceleration which Is 2.45m/s^2 how from here can I get the weight?

 

[azizlwl]

You should not use random value. Just use mg for the weight.
It just a ratio of apparent weight over normal weight.

What is the apparent weight in term of angle.
You can predict when the angle increase, the acceleration increase too.

 

[Stanc]

Ok so how would I take my downward component of 2.45m/s^2 and use itto find weight?

 

[azizlwl]

Its best to substitute the value until the last(if possible).
You should find what is the acceleration in term of the angle 30°.

Then apparent weight Mg-N=ma , N=Mg-ma

The ratio will be Mg-Ma/Mg

 

[Stanc]

wait, does my vertical acceleration look correct? if correct would I just use your equation n-mg= a? But how do I work the angle in??

 

[Stanc]

ya I found the acceleration in terms of the, angle butshouldnt I use the vertical acceleration because the scale on the surface is horizontal?

 

[azizlwl]

You can try. Once you get the answer then check what happen if the angle is 0 and then if the angel is 90°. Thus the value makes sense?

 

[PhanthomJay]

Yes, her downward acceleration is 2.45 m/s^2, very good. Her weight is just mg or 98N if you assume m = 10 kg. Her apparent weight is the scale reading, or the normal force acting perpendicular to the scale. You actually had the correct answer in your post #9, (but your input values were slightly off... ). Using Newton 2 in the vert direction as per azizlw post #11, you get the answer for N. If you had not assumed a mass and instead used her mass as m, you get the normal force directly as a fraction of her weight.

 

[Stanc]

What is wrong with my input values?

 

[azizlwl]

The acceleration down the inclined plane= gSin30°
This acceleration can be resolved into 2 components, down and forward.
Thus downward component= gSin30°Sin30° equal to what you got

Its easier and less prone to calculation error if you only substitute them till last.
Since some of them will cancel out and checking calculation for the last value can be done many times.

 

[Stanc]

Wait so how do I use n- mg = a? Is it n -9.81m= 2.45 which would become 2.45+9.81m = n? or do I isolate m which gives me (2.45+n)/9.81??

 

[azizlwl]

sorry there typo in my last post
N-mg=a should be N-mg=ma

First you have to understand what the equation means.
For a downward acceleration which means the resultant of the forces make the body downward. If we take downward as positive the equation will be
Mg-N=ma. The gravity(Mg) pulls down. The spring(N) of the scale push up and the result the body will accelerate downward(Ma).

N is what the scale is showing the weight while accelerating downward. Thus the ratio is N/Mg

 

[PhanthomJay]

What is wrong with my input values?

I have corrected in red your math in your post #9:

Wait to solve for the fraction of the true weight, would i take acceleration in the y (2.4525m/s^2) and subtract it from gravity so 9.81 - 2.4525 = 7.3575m/s^2 and than multiply that by 10 which gives me [strike]72.2[/strike] 73.575[/color] Newtons. And than would i take [strike]9.81[/strike]10[/color] x 9.81 = [strike]96.2361[/strike]98.1[/color] Newtons and than divide [strike]72.2/96.2361 = roughly[/strike]73.575/98.1 = exactly[/color] 3/4 her true weight[strike]?[/strike].[/color]

 

[azizlwl]

Now without putting any value to m

The ratio= N/mg= (Mg-Ma)/Mg= (g-a)/g
a=gsin30°sin30°

N/mg= (g-g(1/2)(1/2) )/g= 1-1/4= 3/4

\frac {N}{mg}=\frac {g- g.\frac{1}{2}.\frac{1}{2}} {g}=1-\frac{1}{4}=\frac {3}{4}