Stupid physics problem.

[Joe Ross]

Originally Posted by Adam Y.

A window washer pulls herself upward using the bucket-pulley apparatus shown in Fig. 4-40. The mass of the person plus the bucket is 64 kg. The person pulls themselves up the rope on the other side of the pully. Assuming friction doesn't apply. What is the force needed to pull the person up with a constant velocity?

If friction doesn't apply, the person can't grip the rope and can't pull herself up at all.

Originally Posted by russell

I have never taken physics, so feel free to ignore this but I would think that if she was pulling down with force equal to that with which gravity acts on her body then she would not go anywhere at all. She would stay exactly where she was. In order for there to be no acceleration (or constant velocity) all she would have to do is pull with constant force.

That's not correct. Read up on Newton's second law.

[Gary Dillard]

Originally Posted by Joe Ross

If friction doesn't apply, the person can't grip the rope and can't pull herself up at all.

Now THERE'S someone who understands Physics! Reps to you

[UsmanB]

ya i know i probalby did somethign wrong, some how when i did cos 34 i kept getting an negative number :S oh well i was close! im taking my first high school physics class right now

[Joe Ross]

Originally Posted by UsmanB

ya i know i probalby did somethign wrong, some how when i did cos 34 i kept getting an negative number :S oh well i was close! im taking my first high school physics class right now

You probably have your calculator in radians mode and not in degrees mode. A common error.

[Stephen P]

For the box one, remember that

f = uN (N is normal force, perpindicular to the slope, pushing up on box)

To find normal force, resolve the weight (mg) of the box into components perpindicular and parralel to the slope, by taking the sin and cos of the angle.

F = ma = "forwards minus backwards and up minus down"
In this case, the weight vector parallel to the slope is the forward, and is being impeded by the backward force of friction (f = uN).

You might find it handy to know that whatever the mass of an object,
tan(@)= u (u is frictional coefficient, @ is angle of repose, the angle at which something begins sliding)

[UsmanB]

Originally Posted by Joe Ross

You probably have your calculator in radians mode and not in degrees mode. A common error.

yep u were right thought about that but was too lazy to check; checked it now and worked.

[Denman]

heh someone bet me to it ...

The difference is a lot lol....
cos34° = 0.8290375726...
Cos34(rads) = -0.8485702748...


F is actually Frmax remember...

If its in limiting equilibrium, Then the friction is up the slope, and a componant of gravity is down the slope. So along the slope, you get
mgsinα=Frmax
and at a right angle, you have the reaction vs the other componant of the gravitational force. Thus - mgcosα=R
Since Frmax=μR, μ=F/R, so you get mgsinα/mgcosα=μ=tanα
Assuming the object is in Limiting Equilibrium . It will not give you the correct value of μ otherwise it will be lower.


I've attached a little diagram i did in paint.



I am actually studying double maths so i hve done 2 mechanics modules, and doing the final one in a month lol...

[mr_spam]

The entier first problem is also completely ignoring the rotational inertia of the pullies and the streching of the rope that would occur.

[Gasperini]

Right with the friction. My dynamics professor would always trick the class with that. MAXIMUM friction is Mu*Normal force. ACTUAL friction can be less. Remember that when you solve, because it can make equilibrium seem a little wacky. (I'll catch it when I put in numbers and get two answers for the same variable. Watch vector components that appear in both X and Y directions... You can catch errors with them.)