Force and Coefficient of Friction Problem

AI Thread Summary
The discussion revolves around calculating the force of friction and the coefficient of kinetic friction for a girl pushing a light snow shovel at a uniform velocity. The user has calculated the horizontal component of the force as approximately 58 N but is confused about determining the normal force without the shovel's mass. Participants clarify that the normal force is related to the vertical component of the pushing force, which must be considered alongside the horizontal force. The conversation emphasizes that even if the shovel is light, it still exerts a gravitational force, affecting the normal force. Understanding the vertical component of the applied force is crucial for solving the problem accurately.
bijanv
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Okay I'm really stumped on this question and I would be really glad if someone can take me step by step on how to do this question... I have the answer but I have no clue how to obtain it!

The Question is:
A girl pushes a light snow shovel at a uniform velocity across a sidewalk. If the handle of the shovel is inclined at 55 degrees to the horizontal and she pushes along the handle with a force of 100 N, what is the force of friction? What is the coefficient of kinetic friction?

I obtained the answer for the force of friction by just using Fhorizontal = cos55 * 100 = 58 N (answer is 57 N)

Now.. to find the coefficient.. we need to know two things right? The force of friction which we have and the normal force. But how can we know what the normal force is if we are not given mass?



However I'm not sure if that is the correct way to do it... if someone could explain how to get both answers step-by-step it would be great thanks!
 
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Assume a massless shovel (it is, after all, described as light), what would the normal force be?
 
normal force would be 0 then as there is no object for there to be a gravitational force on? but that would make everything (being coefficient of friction and force of friction.. according to the equation Ffriction = mew * Fnormal) equal to 0
 
Normal force would be the ground acting against the shovel, else the shovel would push the ground down.
 
yup but in this case, the normal force is = to the force of gravity on the shovel

if we assume if the shovel is massless... it means there's no force of gravity acting on it and therefore no normal force... correct?

i'm really confused.. so do we have to find the vertical force that she is exerting on the handle?
 
bijanv said:
yup but in this case, the normal force is = to the force of gravity on the shovel
NO, it's not- that's the whole point!

if we assume if the shovel is massless... it means there's no force of gravity acting on it and therefore no normal force... correct?
i'm really confused.. so do we have to find the vertical force that she is exerting on the handle?

You were told the girl pushed the shovel "with a force of 100 N". You correctly (approximately) calculated the horizontal component of force to be "cos55 * 100 = 58 N (answer is 57 N)". What do you think happened to the rest of the force? Where there is a "horizontal component", don't you thing there is likely to be a "vertical component"?

By the way, how did you get "58 N"? What exactly did your calculator say?
 
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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