Sum of the friction and normal forces mechanics problem

AI Thread Summary
The discussion revolves around solving a mechanics problem involving a wedge on a frictionless surface with a block on its slope. The key focus is on determining the maximum and minimum values of the horizontal force F required to keep the block at rest, considering the static friction coefficient μe. Participants clarify that both the normal force and the applied force F contribute to the block's motion due to their horizontal components. The equations derived include the relationship between the forces acting on the block and the wedge, leading to an expression for F in terms of mass m, gravitational acceleration g, and the angle of the wedge. Understanding the roles of these forces is essential for solving the problem accurately.
luiseduardo
Messages
29
Reaction score
0
Hi,

Can anyone explains me how to solve this problem ? I'm very confused what to do.

Homework Statement




"A wedge of mass M subjected to a horizontal force is on a frictionless horizontal surface. Place a block of mass "m" on the sloping surface of the wedge. If the coefficient of static friction between the surfaces of the wedge and the block is "μe", find the maximum and minimum values of force F so that the block remains at rest on the wedge."

[PLAIN]http://img411.imageshack.us/img411/4226/calso.jpg


Homework Equations



Fr = M.a

The Attempt at a Solution



Well, I tried to do, but I have two questions:

What is the force applied to the block that will make it up? It would be a Normal or force F?
 
Last edited by a moderator:
Physics news on Phys.org


Yes, it is the sum of the friction and normal forces that contribute to the motion.
 


Ok, but the force F will contribute to the motion too ?
Or the force F don't acts to the motion only the Normal, is it?
 


Both of them will contribute since they both have horizontal components and the motion is horizontal...
 


Thanks, another question: We need two Normal ?

N (down) = mgcos a - F*sen a
N (up) = F*sen a + mgcos a
 


I found:

[mg(cos a*μe - sen a)] / (sen a*μe + cos a) =< F

F =< [mg(sen a + cos a*μe)] / (- sen a*μe + cos a)
 
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