Center of Gravity: Locate on Uniform Board 20cmx10cm

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To locate the center of gravity of a uniform board measuring 20cm x 10cm with a mass of 200g, two additional masses of 50g and 80g are attached at opposite corners. The calculation involves determining the center of gravity along the x-axis using the formula 20 x 80/(80 + 50), resulting in a position of 12.3cm. For the y-axis, the calculation is 5 x 200/(200 + 130), yielding a position of 3cm. The intersection of these two calculated points provides the new center of gravity for the system. The provided diagrams illustrate the setup and calculations effectively.
holy_kamote
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Center of gravity?

A uniform board 20.0cm x 10.0 cm has a mass of 200g. Masses of 50.0 and 80.0 grams are attached at two corners at the ends of one of the longer sides. Locate the center of gravity.

i have no idea how to solve this help me pls...
 
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you can firstly do it for one direction on xy plane. Firstly work on 50 and 80 for a line from x. then work on 200 and 130 for a line from y. the junction point of these two lines gives you the new center of gravity.
 
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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