2011-03-04, 01:35 AM
There's actually not much physics going on in here. It's just math with certain constants in certain places. Although it's nice to know some physics, knowledge of it isn't completely necessary. Very few actual physics concepts are involved. Namely, the only ones are: sliding friction is proportional to the normal force (force that's exerted on the object by the surface that it's on), force due to gravity is mass*(acceleration due to gravity), and work done is equal to the force times the displacement (and that these two must be in the same direction). If you're unfamiliar with the symbols, here they are:
m = mass
g = acceleration due to gravity
d = distance
F = force = mass*acceleration
W = work = force*distance
μ = coefficient of friction
These are really just constants. It might be nice to already know some of these, but the important ones should have already been given to you in your math course.
The first problem is entirely geometric. Once you know that the force downward is the mass, m, times the acceleration due to gravity, g, and that it's attached to two cables from the ceiling, you can automatically say that each wire will be exerting the same amount of upward force to hold up the mass. You know the y-component and you know the ratios between each of the sides of the triangles. You can rescale the force from the vertical direction to the final direction.
For the second problem, if you want a geometrical interpretation, this is possible from the picture. If you don't know any calculus, you can just skip to the last two lines. What's necessary here is that you know the work is defined as the displacement times the force that's acting in the same direction as that displacement.
For the third problem, I pretty much made the answer very simple to see. The force done is done at an angle, but the box is moving horizontally. Since you only want to consider the component of the force that's in the same direction as the motion of the box. It's very easy to see that this is just (F1)*cos(θ
and (F2)*cos(φ, and since work is force times distance, then you can easily find that the total work done is just those two added up times the displacement.
The following shouldn't be new to you, but I will list them anyways:
r = parametrization of the curve
In other words, you can break down any function into an x path and a y path.
For example, take f(x) = x^2. You can rewrite that as:
f(t) = <t,t^2>, or ti + t^2j. That means that for every t step you take outward in the x direction (the first half), you take t^2 steps in the y direction (the second half).
φ and θ are angles
i, j, and k (with the ^ hats) are the unit vectors, meaning a vector with lengths of 1 in x, y, and z directions, respectively.
Hopefully, this wasn't too abstract. If you didn't understand any of this at all, at least you should know how to plug variables into the solutions and see if you obtain the same results.
m = mass
g = acceleration due to gravity
d = distance
F = force = mass*acceleration
W = work = force*distance
μ = coefficient of friction
These are really just constants. It might be nice to already know some of these, but the important ones should have already been given to you in your math course.
The first problem is entirely geometric. Once you know that the force downward is the mass, m, times the acceleration due to gravity, g, and that it's attached to two cables from the ceiling, you can automatically say that each wire will be exerting the same amount of upward force to hold up the mass. You know the y-component and you know the ratios between each of the sides of the triangles. You can rescale the force from the vertical direction to the final direction.
For the second problem, if you want a geometrical interpretation, this is possible from the picture. If you don't know any calculus, you can just skip to the last two lines. What's necessary here is that you know the work is defined as the displacement times the force that's acting in the same direction as that displacement.
For the third problem, I pretty much made the answer very simple to see. The force done is done at an angle, but the box is moving horizontally. Since you only want to consider the component of the force that's in the same direction as the motion of the box. It's very easy to see that this is just (F1)*cos(θ
and (F2)*cos(φ, and since work is force times distance, then you can easily find that the total work done is just those two added up times the displacement.The following shouldn't be new to you, but I will list them anyways:
r = parametrization of the curve
In other words, you can break down any function into an x path and a y path.
For example, take f(x) = x^2. You can rewrite that as:
f(t) = <t,t^2>, or ti + t^2j. That means that for every t step you take outward in the x direction (the first half), you take t^2 steps in the y direction (the second half).
φ and θ are angles
i, j, and k (with the ^ hats) are the unit vectors, meaning a vector with lengths of 1 in x, y, and z directions, respectively.
Hopefully, this wasn't too abstract. If you didn't understand any of this at all, at least you should know how to plug variables into the solutions and see if you obtain the same results.