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 she increases this acceleration by 14% how much does she accelerate. I hate my physics class. This is the only class where I do my homework online. Get the answer wrong on one part of the question, use the wrong answer for another part, and get the right answer for another part. Here is another one:
A 19.0 kg box is released on a 34.0° incline and accelerates down the incline at 0.03 m/s2. Find the friction force impeding its motion. What is Mu?
I wouldn’t be asking these questions if the didn’t confuse me and my teachers confuse me even more.
Adam,
My apologies for what I am about to say… These are fairly simple questions… If they confuse your teachers, find a new teacher.
JMHO
No. Im sorry. I meant to say there replies confused me. The one person said he would have to see my work to figure out how I got the right answer for the one part and the other one tried to help me but they just confused me even more. I honestly to God went to extra help for both of these problems. The graduate student was just as confused as I am. She was a very helpful though on other problems I had “duh” moments on. I thought the answer would be 64kg * 9.8m/s2 first problem but nooo. That isn’t the right answer. Please answer this also. If you get the first part wrong then you should get the wrong answer for the second part also. Right?? Im asking the last bastion of nerds that I know. Please help me…
i did a quick search of ur problem on google and found this problem on a forum:
A window washer pulls herself upward using the bucket-pulley apparatus show in Fig. 4-42 (the woman is in a bucket, a rope tied to the top of the bucket, and she’s pulling on the downward direction of rope that goes up over the pulley) (a) How hard must she pull downward to raise herself slowy at a constant speed? (b) If she increases this force by 10 percent, what will her acceleration be? The mass of the person plus the bucket is 65 kg.
This was the reply from a member.
Find the accelerations of all four parts. (g=10)
a = (m2-m1)/(m1+m2)]g
a1 = (180kg - 120kg)/(180kg + 120kg)]10m/s^2
a1 = 2m/s^2
a2 = (120kg - 80kg)/(120kg + 80kg)]10m/s^2
a2 = 2m/s^2
a3 = (80kg - 40kg)/(80kg + 40kg)]10m/s^2
a3 = 3.33m/s^2
a4 = (40kg - 0kg)/(40kg + 0kg)]10m/s^2
a4 = 10m/s^2
Find the time in seconds.
s = .5at^2
16m = .5 (2m/s^2) t^2
t^2 = 16m
t1 = 4s
t2 = 4s (same acceleration)
16 = .5 (3.33m/s^2) t^2
t3 = 3.1s
16 = .5 (10m/s^2) t^2
16 = 5t^2
t^2 = 3.2
t4 = 1.8s
Adding the times of 4+4+3.1+1.8 = 12.9 seconds.
And for your other problem i found this on google. http://faculty.salisbury.edu/~jwhoward/Physics121/html/lecture4.htm
that is the exact way to do ur problem, scroll down and u’ll find a problem that reads “An 18.0 kg box is released on a 37.0° incline and accelerates down the incline at 0.270 m/s2. Find the coefficient of friction and the frictional force.” and they have a picture and show u how to solve it 
good luck and im sure that if u looked in the chapter ur doing right now u could find perhaps an example, i know the physics book i have right now doesnt have each example for each type of problem so u have to play with equations given.
Acckkk…
You just gave me the answer to the problem above that post. Nice try but wrong problem.
good luck and im sure that if u looked in the chapter ur doing right now u could find perhaps an example, i know the physics book i have right now doesnt have each example for each type of problem so u have to play with equations given.
I did play with the equations. The only problem is that if Im getting the wrong answer for one part I should be getting wrong answers both parts. It’s double jeporday. Get one part worng and get all the other parts wrong. Friction=Mu*FN. Well how do I get the right answer when my force of friction is wrong.
i thought perhaps u could see waht they did and perhaps it could help u 
umm here ill actually try sovling the rpoblem for u if i can. i didnt even try before.
btw the picture for that pulley problem, is there any other info in it?
Nope. She is pulling herself in a bucket attached to a pulley. Logically here is all the information that I used. Constant velocity means no acceleration. Thus this means that there is no net force being acted upon it. That means that she must pull down with the same force that she weighs.
i did it and got -.620 for the mu. and the force in x and y direction are 98.1 and -158.17
I did the second problem and got the force of friction = 103.55 N, and mu = 0.67.
First, you find the weight of the box.
w=mg=(19)(9.8)=186.2 N
Then you find the parallel force and the normal force.
Fp=wsin(angle)=186.2sin(34)=104.12
Fn=wcos(angle)=186.2cos(34)=154.37
So by F=ma:
Fp - F(friction) = ma
104.12 - F(friction) = (19)(0.03)
F(friction)=103.55 N
Now we can solve for mu:
F(friction)=muFn
103.55=mu154.37
mu=0.67
Does this help at all?
Constant velocity means no acceleration. Thus this means that there is no net force being acted upon it. That means that she must pull down with the same force that she weighs.
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.
I don’t know what you’ve already covered in Physics, but before I answer your problems you’ll have to indulge me in a story…
I took Physics my freshman year of college, and we had 3 exams which were all worth 33 points. The class average on each one was about 12; the second highest score on each one was 18-19; one kid got a 33 on the first test and a 32 on the second; he didn’t have to take the final. I remember looking at the problems in my Physics book with a blank dumb stare - how were we expected to understand this stuff?
Fast forward 3 years. I walked into a room of a freshman, he had the exact same Physics text book and the exact same blank dumb stare on his face. I looked at what he was doing, showed him how to solve the problems in about 5 minutes, and he thought I was some kind of genius. That’s when I realized that it just takes alot of practice and exposure to different methods for it to sink in. We covered the same material in Physics, then Statics, then Dynamics, then strength of materials, then machine design, then mechanics - when you use the same stuff in different applications somewhere along the line it sinks in. I’m not sure when it did for me, but it did. So don’t give up.
Anyway, for both of these problems you need to draw a free body diagram, which shows all of the forces acting on a system or body. If they equal zero, you have a statics or constant velocity problem. If they don’t equal zero, you have a dynamics problem.
Number 1a: constant velocity means forces up equals forces down.
Total force acting down = mg
Total forces acting up (I’m assuming there’s a single pulley at the top) equals F {the tension in the rope pulling on the bucket} + F {the tension in the rope pulling on your hands, which equals the force you are pulling on the rope}. Note that the tension along any point on the rope is equal to F. Therefore, 2F=mg, F=mg/2 = (64)9.81/2 = 314 Newtons
Number 1b: I’m guessing you meant increase force by 14%, but it doesn’t matter.
Total force acting down = mg
Total forces acting up = 2F1.14 = mg1.14
Difference = mg(1.14-1) = mg.14 = net force
Since F=ma, ma=mg.14, or a = .14g or 1.373 m/s^2
Number 2: sum forces normal to the plane
F = mgcos(34)
sum forces parallel to the plane
mgsin(34) - (mu)(F) = ma, or
mgsin(34) - (mu)mgcos(34) = ma
The mass term falls out
[9.81sin(34)-(.03)]/[9.81cos(34)] = .67 = mu
Adam,
Although I would have preferred that you solve these problems on your own (and congrats to the physics majors out there who divined the answers), you should note the following:
In both answers, the second part is really independent of the first. Example: plug a different weight into the second problem and Mu is exactly the same. the block could have a mass of a billion kilos with exactly the same answer, Mu = 0.671! In the same way, the answer to problem 1b is dependent only on the acceleration due to gravity.
What you highlighted as a point of frustration is one of the beauties of the universe… Cool eh?
ey?
mu < 0 ? lol that implies the friction on the slope is accelerating object down the slope, and friction acts against the slope…
If friction doesn’t apply, the person can’t grip the rope and can’t pull herself up at all.
That’s not correct. Read up on Newton’s second law.
Now THERE’S someone who understands Physics! Reps to you
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.
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)
yep u were right thought about that but was too lazy to check; checked it now and worked.
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.
:ahh:
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…