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Consider a gas spring in the vertical 90 degree position and a 275lb weight sitting on the top of it. There is a downward force canceling out an upward force. Now consider we slide out the base 1 degree assume the shock cannot fall over.

Using Trigonometry I calculate using the sine function I use a right triangle configuration with 275 being my Y component force and solving the sine equation I calculate (R)Hypotenuse as being 275/(Sin89) I get 275.042=R.

My question is why is the force calculated greater? I would think the y-component force against the weight will now be less than 275 (275-.042) and the R-component force would be 275.042-275=.042.

More information I probably didn't do a good job of setting up this question. Imagine an automotive floor jack holding a 275lb weight. now replace the hydraulic piston with a gas spring and lets say the spring is at 90 degrees to the floor to start. At this point the spring is supporting the weight and we are in balance. I want to calculate the force on the spring as the bottom of the spring is pivoted outward from under the weight in one degree increments. Essentially the spring is moved from a vertical position to a horizontal position. Each time the spring is moved it is replaced with a spring that has enough force to keep the weight balanced. My thinking is that the spring will be 275 on vertical and zero when it is laying down and the weight is on the floor. In summary I was trying to come up with a formula to plug into an excel spread sheet to show the amount of force on the spring at each degree of increment. Hope this helps.

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Force diagram

Consider the force exerted by the gas spring, $\vec{F}$, to be the vector sum of the horizontal component $\vec{F}_x$ and the vertical component $\vec{F}_y$, as shown above:

$$\vec{F}=\vec{F}_x+\vec{F}_y$$

Using trigonometry, the scalars (magnitudes) become:

$$F_x=F\cos\alpha$$ $$F_y=F\sin\alpha$$ And with Pythagoras:

$$\begin{align}F &=\sqrt{F_x^2+F_y^2}\\ &=\sqrt{F^2\cos^2\alpha+F^2\sin^2\alpha}\\ &=\sqrt{F^2(\cos^2\alpha+\sin^2\alpha)}\\&=F\end{align}$$

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You are right in thinking that the y-component force will be smaller. 275 is not the y-component when the spring is tilted, it is the hypotenuse.

The weight will always have the same force pulling it down, which is the gravitational pull $F_{g} = mg$. If initially the gas spring is not tilted and the weight remains on top of it, that means that there is an upward force coming from the spring which compensates the downward force, and so they have the same magnitude

$F_{s} = F_{g} = 275$ lbs

(I'm assuming 275 lbs is the force, not the mass, but it works the same if you multiply by g)

Then, when you tilt the spring, if you do not change anything else it will still push with the same force, but the direction of $F_s$ will have changed.

To figure out the new upwards force you need to decompose $F_s$ into its x- and y-components. Take $F_s$ as your hypotenuse, and you will have

$F_y = 275 \times \sin (89) \approx 274.96$

$F_x = 275 \times \cos(89) \approx 4.8$

So, the weight will fall down and to the side (which depends on which way you tilted the spring).

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  • $\begingroup$ Thank-you..but the weight will always be straight down 275 now matter how the spring is tilted. what am I missing? $\endgroup$ – Son of Fire Nov 15 '16 at 21:23
  • $\begingroup$ What exactly are you trying to calculate? Are you trying to make it so that the weight doesn't fall? I don't think that can happen because even if the force of the tilted gas spring has a y-component that compensates for the downward gravitational pull of the weight, it will have an x-component that will push it to the side and then it will fall. $\endgroup$ – DarthZapps Nov 17 '16 at 3:17

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