Concurent and Parallel Forces 2

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Four forces, each with a magnitude of 2750 lb, act at the same point with 30-degree angles between adjacent forces. The calculation for the resultant force involves breaking down each force into its x and y components. The initial claim of the resultant force being zero in both x and y directions is incorrect. Proper vector addition shows that the resultant force is not zero, indicating a miscalculation in the original analysis. Accurate vector resolution and summation are essential for determining the correct resultant force.
kgbwolf
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Four forces each of magnitude 2750lb act at the same point. The angle between adjacent forces is 30degrees. find the resultant force. I worked it out and got FR= 0 in the x and 0 in the y is that right??
 
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Let's take F1 as X-direction. Then F2 makes 30deg with it and similarly F3 makes 60deg while F4 makes 90 deg or denotes the Y-direction.
Now add the vectors properly - 2750x + 1375*root3x + 1375y + 1375x + 1375*root3y + 2750y. This isn't anyway 0. Why did you get them wrong?
 
i see what ur saying thanks
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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