Raising Barrel w/ Pipes: Min Force Required | Physics Homework

  • Thread starter Thread starter animedragon
  • Start date Start date
  • Tags Tags
    Pipes
AI Thread Summary
To determine the minimum horizontal force required to raise a 100 kg barrel using pipes, the forces acting on the barrel must be analyzed using Newton's laws. The contact forces from the pipes on the barrel are both horizontal and vertical, necessitating a trigonometric approach to resolve these components. By drawing lines from the barrel's center to the pipes and calculating the angles involved, the vertical force needed to lift the barrel can be derived. This vertical force must equal the weight of the barrel, which is 980 N (100 kg x 9.8 m/s²). Solving for the horizontal force applied through the pipes will yield the required minimum force to raise the barrel.
animedragon
Messages
1
Reaction score
0

Homework Statement



Michael is trying to raise a barrel using sections of pipe.The three objects are horizontal cylinders resting on level ground. The barrel has a mass of 100 kg and its diameter is 2.0m. Each pipe has a diameter of 0.50 m. What minimum horizontal force must his feet exert on the pipe to raise barrel from the ground? Neglect Friction.


Homework Equations



Newton's first law, Newton's second law, Newton's third law...

The Attempt at a Solution



well, i figure that the pipe in the corner against the wall has no acceleration, but because of Newton's third law exerts a force equal to force applied onto the barrel (force contact). now, I am just confused after where to go now.

i thought about it, and the distance between the two cylinders, pipe and barrel, must 1.5m.

thats as far as i got. PLEASE HELP! D:
 

Attachments

Physics news on Phys.org
Welcome to PF, dragon.
I suggest you draw a line from the center of the barrel to the center of each pipe, passing through the points where the pipes contact the barrel. The force of each pipe on the barrel will be along those two lines and will be partly horizontal and partly vertical. Draw the component of the applied force vector that is along the contact line and calculate its portion of the applied F using trigonometry (you will need the angle of the contact line above horizontal). Then multiply by the appropriate factor to get the vertical component of that force. Double it because you have an equal force from the other side. That should give you a number time F, which is equal to the weight of the barrel. Solve for F.
 
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...
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}...
Back
Top