Forces applied to a spring-loaded gas pedal

[Theexploer]

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.

 

[BvU]

Hello @Theexploer ,

$\qquad$ !​

So this is a $G$ ,

right ?

But who is $T$ ? And $P$ ?

And did you render the complete problem statement ?

$\$

 

[Theexploer]

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.

 

[BvU]

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

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):

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

$\$

 

[haruspex]

T=4.7N

Looks right.
But I don't understand the second question either. Is it a translation? Is R defined anywhere?

 

[TSny]

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.

 

[haruspex]

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.

 

[Theexploer]

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