Stupid physics problem.

[UsmanB]

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/~jwhowa...l/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.

[Adam Y.]

Originally Posted by UsmanB

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.

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.

[UsmanB]

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?

[Adam Y.]

Originally Posted by UsmanB

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.

[UsmanB]

i did it and got -.620 for the mu. and the force in x and y direction are 98.1 and -158.17

[Pat Fairbank]

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=w*sin(angle)=186.2*sin(34)=104.12
Fn=w*cos(angle)=186.2*cos(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)=mu*Fn
103.55=mu*154.37
mu=0.67

Does this help at all?

[russell]

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.

[Gary Dillard]

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 = m*g
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, 2*F=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 = m*g
Total forces acting up = 2*F*1.14 = m*g*1.14
Difference = m*g*(1.14-1) = m*g*.14 = net force
Since F=m*a, m*a=m*g*.14, or a = .14g or 1.373 m/s^2

Number 2: sum forces normal to the plane
F = m*g*cos(34)
sum forces parallel to the plane
m*g*sin(34) - (mu)*(F) = m*a, or
m*g*sin(34) - (mu)*m*g*cos(34) = m*a
The mass term falls out
[9.81*sin(34)-(.03)]/[9.81*cos(34)] = .67 = mu

[Denman]

Originally Posted by UsmanB

i did it and got -.620 for the mu. and the force in x and y direction are 98.1 and -158.17

ey?
mu < 0 ? lol that implies the friction on the slope is accelerating object down the slope, and friction acts against the slope...