2007-10-29

Today's nonsense
- No Dakota.
- It's a setup- they're all Twix.
- Balto the movie?
- Oil content of the cheez-it bag (Katie). Is it intimidating?
- Spiller likes to mock me typing.
- Coach Burnett came in yesterday and said that Spiller had messages on the phone because of the red light, and Burnett went through and explained it, and how come when you call in to the school and try to dial the extension it doesn't let you? Maybe to prevent interruption of the class. But what about on the weekends?
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There will be a take-home physics quiz because the grades on the test were pathetic.

Free-body diagrams

Be neat. Free-body diagram- a diagram that shows all the forces acting on a system.

ex 1) disguised, you enter a dogsled race. You're disguised and dressed up as a dog. You're catching on, you can do this. Excuse the excitement. You pull on a rope attached to the sled with a force of 150 N at 25 degrees w/ the horizontal. So you will have to realize that the larger the angle is with the horizontal, that you're wasting a lot of force, and the only force causing it to move is the horizontal force, and the vertical force is trying to lift the sled. The mass of the sled is 80 kg and there is negligble friction. A) Draw a free-body diagram for this.

B) Find the acceleration of the sled. The y-acceleration is going to be zero, unless you're traveling over a frozen lake that's not so frozen, or you've just gone over a hill. You can use the forces and the masses (as in the diagram) to solve for the horizontal acceleration. In this case we do not have to deal with vertical acceleration, so there will be no 'triangle for acceleration'. Find a_x. So we can use Tcos25 = m(a_{_x}). So when you pull with a force of 150 N, the force is going to have that tension.

a_x = 1.699 m/sec^2

C) Find the normal force, F_{_n}, exherted by the surface on the sled. The rope is pulling up on the sled as well- the rope is not horizontal. We said that the sum of the forces in the y-direction was zero for the rope. We have 150sin25 + F_{_n} - 80(9.81) =0 and somebody will realize that "g" is going to be 9.81 and that's the acceleration due to gravity and we are not doing -9.81 because we are taking into account that the weight is going down, and that's why we said -80(9.81) in that line of equations from just a few moments ago. F_{_n} = 721.407 N. Now, do you understand, well, actually, you might not understand, but I may as well ask the question, and make you think about it, did you figure out what you did wrong, okay, Brooke is still writing if you're still wondering, I shouldn't have to single out anybody, and we're still on the problem, D, so, it could be rounding up but that would be rounding but, so [pause].

D) What is the greatest tension that can be applied to the rope without lifting the sled off the surface? Now if I was to pull with a force great enough, I could just pull it off the ground, so the question is going to be what is going to be the greatest tension that can be applied without lifting it off the ground, and the greater the tension in the rope, the more vertical force as well as. So we want the vertical to always be zero.

F_{_n} > = 0
You want your normal force to be greater than or equal to zero. So let's look at the worse-case scenario. Take your sum of y-forces, solve for F_n, move your stuff to the other side, and you'll have F_{_n} = mg - Tsin25 and then you plug in numbers, setting F_{_n} to zero = (80)(9.81) - Tsin25. Then solve and get T = 1856.995 N. Anything greater than that and you'll be lifting the sled off the ground.

2) You unload a truck by sliding its cargo down, so it is going to have a ramp, down a ramp, with roller, assume the ramp is frictionless. The ramp is inclined at an angle theta. For a box of mass m find the acceleration of the box as it slides down the ramp and the normal force exerted by the ramp on the box. The normal force is perpendicular. So in this case the normal force is off to an angle to the typical surface that we associate with.

The normal force is straight in the y-direction, but the weight is at an angle in the problem (but universally, the weight is as it usually is). So from this we are going to get our horizontal and vertical components. So let's look at the sum of the forces now. Shall we? We shall. So, what about our, horizontal forces? To find F_x it's going to be SOHCAHTOA and it's going to be sine. So we're looking for the one that's opposite, we have a parallel line below our dotted line that parallels the ramp. So in this case the sum of F_x is going to be wsin(theta) and then any other horizontal forces. It's just like as if we had a vector in the fourth quadrant. So, what's the net force in the y direction? It's mass times the a_y acceleration, but is this thing accelerating in the y-direction? So F_y should be zero.

Since F_y = 0, we know the mass is m, we know g, we know theta, so what is the normal force?
F_{_n} = (mg)cos(theta)
And what's the acceleration? a_{_x} = gsin(theta) because the masses just cancel themselves out via division.

Where would friction come into play? In the sum of F_x and so on we would have other stuff to do.