Forces applied to a spring-loaded gas pedal

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  • #1
Theexploer
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TL;DR Summary: An accelerator pedal is located steadily on our line at point O, the spring AB is perpendicular to the accelerator pedal, keeping it balanced at an angle of 45 degrees. a = 45* the weight of the accelerator pedal is 10N and it is loaded on G. OG = 10 cm OB = 15 cm.
Question = Calculate the intensity of the force T applies to the accelerator pedal.
Question = Determine the direction, side, and density of R from point O.

For the firtst question i did like that =
As the pedal is in balance the sum of the moment of
external forces are zero
M(P)+M(T)+M(R)=0
M(R)=0 because it meets the axis of rotation
So M(P)+M(T)=0
If we choose a positive direction towards P
M(P)=P.dp or dp = OG.cosalpha=OGcos45°
M(T)=-T.OB
T=OGxcos45°xP/OB=10x10xcos45°/15
T=4.7N

But i'm not sure sor i can't do the second one.
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  • #2
Hello @Theexploer ,
:welcome: ##\qquad## !​

So this is a ##G## ,
1707945734347.png
right ? :wink:

But who is ##T## ? And ##P## ?

And did you render the complete problem statement ?

##\ ##
 
  • #3
Yes, it's a G wich is the point of gravity, T is the tension and P is the Gas Pedal wich have a mass of 10N.

so i rendered the complete problem statement.
 
  • #4
Theexploer said:
T=OGxcos45°xP/OB=10x10xcos45°/15
Let me asssume this translates as follows:
## OG## is a given distance, 0.1 m
x is 'multiply'
##P## is not the gas pedal but the weight of the gas pedal, 10 N

Theexploer said:
we choose a positive direction towards P
Does that mean ##P## = +10 N ? (i.e. down is positive) ?


##OG \cos 45^\circ## is the perpendicular distance of ##P## wrt ##O##
The 45 degrees is the angle AOB (not the other one, BAO)

##OG \cos 45^\circ P## is the torque you call M(P)
it's acting in clockwise direction (which you consider positive, right ?)

and M(T) is the torque the spring exerts. It's acting in counter-clockwise direction (so ##T## is pointing up?)
It is pointing perpendicular to OB, so the torque is ##M(T) = -T \ \text{OB}##

And the torque balance ##M(P)+M(T)=0## yields ##T= OG \cos 45^\circ P/ \text{OB}##

In a more conventional notation (torque as a vector and with y+ = up):

1707956983913.png


##\vec \tau_\text {left} +\vec \tau_\text {right} = 0 \Rightarrow ## ## \vec{\text {OG}} \times \vec mg ## ##+ \vec {\text {OB}} \times \vec T = 0 \Rightarrow## ## |OG|\, mg\sin\theta_1+ |OB|\, T\sin\theta_2 =0 ##
[edit] fixed typo
And I confess I have no idea what is asked in part 2. 'Density of ##R##' ? R doesn't occur in the story ...

##\ ##
 
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  • #5
Theexploer said:
T=4.7N
Looks right.
But I don't understand the second question either. Is it a translation? Is R defined anywhere?
 
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  • #6
I think R is the reaction force acting on the pedal at the pivot O. They want the magnitude ("density"?) and direction of this force. I don't have any idea what "side" of R means. The wording does appear to be a translation.
 
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  • #7
TSny said:
I don't have any idea what "side" of R means.
Maybe "direction, side, and density" means angle to the horizontal, to the left or to the right, and magnitude.
 
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  • #8
BvU said:
Let me asssume this translates as follows:
## OG## is a given distance, 0.1 m
x is 'multiply'
##P## is not the gas pedal but the weight of the gas pedal, 10 N


Does that mean ##P## = +10 N ? (i.e. down is positive) ?


##OG \cos 45^\circ## is the perpendicular distance of ##P## wrt ##O##
The 45 degrees is the angle AOB (not the other one, BAO)

##OG \cos 45^\circ P## is the torque you call M(P)
it's acting in clockwise direction (which you consider positive, right ?)

and M(T) is the torque the spring exerts. It's acting in counter-clockwise direction (so ##T## is pointing up?)
It is pointing perpendicular to OB, so the torque is ##M(T) = -T \ \text{OB}##

And the torque balance ##M(P)+M(T)=0## yields ##T= OG \cos 45^\circ P/ \text{OB}##

In a more conventional notation (torque as a vector and with y+ = up):

View attachment 340350

##\vec \tau_\text {left} +\vec \tau_\text {right} = 0 \Rightarrow ## ## \vec{\text {OG}} \times \vec mg ## ##+ \vec {\text {OG}} \times \vec T = 0 \Rightarrow## ## |OG|\, mg\sin\theta_1+ |OB|\, T\sin\theta_2 =0 ##

And I confess I have no idea what is asked in part 2. 'Density of ##R##' ? R doesn't occur in the story ...

##\ ##
Thanks for your help
 
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