What Went Wrong in My Incline Friction Problem?

AI Thread Summary
The discussion focuses on a physics problem involving incline friction, where the original poster struggles to identify an error in their solution. Key feedback indicates a misunderstanding regarding the force of the rope, which was incorrectly assumed to equal the weight of the hanging mass. Instead, the correct approach involves defining the rope tension as T and applying Newton's second law to both masses. By combining the resulting equations, one can accurately solve for the acceleration. This clarification is essential for resolving the problem effectively.
houseguest
Messages
16
Reaction score
0
Hello, this is a physics problem I worked out an cannot determine what I did wrong.
Attached is the problem and my solution. Any help would be very appreciated.
Thanks.

Homework Statement


Attached

Homework Equations



F = ma, f_k = Mu_k * N, dx = v0t + 1/2at^2

The Attempt at a Solution


Attached
 

Attachments

  • prb1.JPG
    prb1.JPG
    44.4 KB · Views: 497
Physics news on Phys.org
Looks to me that you assumed the force of the rope on the mass was equal to the weight of the hanging mass. That's not true. (If it were true, the hanging mass would be in equilibrium.)

Instead, call the rope tension T. Apply Newton's 2nd law to both masses, then combine the equations to solve for the acceleration.
 
Ooooh yeah. Thanks!
 
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}...
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...
Back
Top