Physics Help: Lifting a 200 lb Beam w/ Unequal Weight

Thread starter NoBodyKnows
Start date Nov 15, 2009
Tags

Lifting Physics

AI Thread Summary

To lift a 200 lb beam that is 20 feet long with one man bearing 50% more weight, the stronger man should carry 120 lbs while the weaker man carries 80 lbs. To achieve this balance, the stronger man needs to position himself closer to the heavier end of the beam. The calculations involve determining the appropriate distances from each end of the beam based on their respective weights. The discussion emphasizes the importance of understanding torque and balance in this scenario. The problem can be solved with basic physics principles regarding weight distribution.

Nov 15, 2009

NoBodyKnows

Not sure about this one, any help?

Two men want to carry a wooden beam weighing 200 lbs which is 20 feet long. Both men are equal height however one is stronger than the other and wishes to bear 50% more the weight. How far from the end of the beam should both of the men be to make this possible?

Physics news on Phys.org

Nov 15, 2009

Chi Meson

Science Advisor

Homework Helper

That is a very easy problem. Do you really have no idea how to start? Let's see what you think first.

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

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}...

View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

View full post »