# Easiest way to pull an object

1. Jul 1, 2009

### Joans

I found interesting to me this topic and tryed to analyze.
http://img505.imageshack.us/img505/6655/49209931.jpg [Broken]

At which angle it is eseaiest to pull an object?
I got equation
$$F=\frac{\mu mg}{cos(\alpha)(\mu tan(\alpha)-1)}$$ or $$F=\frac{\mu mg}{\mu sin(\alpha)+cos(\alpha)}$$

I dunno this math very well unfortunately, but I would be interested to see a plot:
how mostly optimum angle is dependent from mu, and for example then mu is 1 alpha is 45, and so on...

heh sorry for the paint and spelling...

Last edited by a moderator: May 4, 2017
2. Jul 1, 2009

### jimcharoen

Re: Easyest way to pull an object

It depends on mu actually. Think of 2 extreme cases. First, mu is 1 (friction is extremely high). In which case X = Y, the answer is 45 degree as you said. But if mu is 0 (frictionless), the answer is clearly 0 degree (the force required to move the object is close to zero).

I'm not sure though; ha haaa

Mr Peetiya

3. Jul 2, 2009

### Brin

Re: Easyest way to pull an object

You're looking for a maximum. Use your calculus brain. How does one find the maximum or minimum of a curve?

4. Jul 2, 2009

### Joans

Re: Easyest way to pull an object

I would use calculus if i know how to use it, in school i do not have lessons with it, unfortunately, since I am 11grader. But infact i know quite a lot about it. But still I dunno how to found derirative of the bottom. $$f(\alpha)=\mu sin(\alpha)+cos(\alpha) f'(\alpha)=\mu cos(\alpha)-sin(\alpha)$$ ?? When to make it to zero and solve it? How to solve what equation when? It's homogenic .. divide by cos alpha and whola? :)
And does best angle depends from $$\mu$$ lineraly? In fact this topic is quite clear, just math's is not very clear.

5. Jul 2, 2009

### Brin

Re: Easyest way to pull an object

Well, you seemed to have gleaned the important part of the derivation anyways.

$$f(\alpha) = \frac{\mu mg}{\mu sin(\alpha) + cos (\alpha)}$$

$$f'(\alpha) = \frac{\mu mg (\mu cos(\alpha) - sin(\alpha))} {(\mu sin(\alpha) + cos(\alpha))^2} = 0$$

because $$\mu mg$$ is constant, and the denominator can't be zero, we can simplify this problem a bit by focusing on the only part that can be zero.

So, you see why I thought you did well on calculating the f' you did:
$$\mu cos(\alpha) - sin(\alpha) = 0$$

Then

$$\mu cos(\alpha) = sin (\alpha)$$

So,
$$tan(\alpha) = \mu$$

Then arc tan both sides to get an explicit value for alpha. By analyzing this function, e.g. you can see that if there is no friction mu = 0, the best pull is the directly horizontal pull (i.e. alpha = 0). If you have mu = 1.0 the best pull is at alpha = 45 degrees.

If you don't know calculus, this problem probably seems a little out of your league. But I am fairly confident there is a geometrical approach as well, that is within your limits if you're an intelligent high school student, or a bored undergrad. If you have the time, and are still curious, I'd recommend seeking out that way.

6. Jul 3, 2009

### Joans

Great, thanks!