Question

Problems with classic mechanics (with calculus)

Does any of you have good physics books that clearly explains the process of various physics problems? especially the newton laws and 3 dimensional motion of objects, I find that I can overcome that by simply going into problems and experimenting with them, sadly I end up taking 30 minutes or so solving one problem, I'm not sure if i'm too slow or being distracted but that remains as is and I want to become smoother and more flexible when it comes to physics problem from various views, as of now I only see my self having trouble with tension.

Answer 1

It sounds like you think you take too long to solve a problem and that all problems can be solved quickly. I would have to see the problem to judge that. Frankly some problems do take a good amount of time to solve. Often it is in determining the correct approach and setting it up and sometime it in the solution of the equation that takes time. I don't know of any text that can help alleviate all the work all the time.

Answer 2

Physics can be quite challenging to master - many years of work to be a physicist, don't be too hard on yourself : )

Answer 3

@ProfBrainstorm thanks for the encouragement, but I still wish to set my tactics for learning right... also, @gleem, I realize that some problems do take time due to their long solutions or puzzling approaches... It's just...

let's take problems like evaluating a force in an incline, let's say there's a particle on an incline that's phi degrees and has an F force parallel to the incline. I was told to find said force I was given 2 other forces to play my problem with, Weight. and Normal force, to find said Force, I am to consider that it, and the anti(negative) normal force are components of Weight?... That said, It wasn't explained as to why.

I did deduce that, the normal force and its perpendicular F force are both systemically pushing the particle object down the incline, which is a reason why they are components of the Weight Force, that said I still wasn't satisfied, I don't feel like I realize the whole mathematical, Or perhaps experimental reasoning as to why these 2 forces are components of Weight. I just realized I have problems like that with my reasoning and wish if anyone had tips as to how to clear them.

Sorry this is long but that's all! thanks to both of you, anyways!

Answer 4

Any force can be "decomposed" into any number of component forces where the vector sum of these components add up to the original force.. We usually consider using two component particularly two that are at right angles to one another.. In the inclined plane problem the applied force is the weight of the object on the plane. It is directed down.
Now we know from experience that there is a force directed parallel to the plane and driving the object down the plane. We also know that the object is pressing against the plane. If the plane is frictionless this component is irrelevant. But if there is friction we know that the force of friction opposes the driving force and must be subtracted from it. to get the net effective force.

I must leave for a while but I will return.

Answer 5

Not a book, but I know HyperPhysics is really good.

Answer 6

Sorry I took so long. let me continue.

The force of friction is of course along the plane in the opposite direction of the component causing the body to slip down the plane. But note that the direction of the component responsible for the friction is not in itself useful. Only the magnitude in this case is useful.

In you have a force or any vector V and you want to know its magnitude in any direction D that is the component of V in the direction D multiply V by the cosine of the angle between V and D. You choose D for what ever reason with V usually given although sometime it may have to be derived separately .

Answer 7

Let me show you another problem resolving vectors. You have a weight of mass m suspended in the middle of a cable attached to two walls. You want to find the tension in the cables the tension is supporting the weight. See the attached diagram.

The tension in each side of the cable is the same by the symmetry of the problem. It is the component of the tension in both parts of the cable that support the weight. so you want the components (2) of the tensions in the direction opposite of the force of the weight.
As you can see if a is close to 90 deg the tensions can be quite large for relative small masses.

If the weight where suspended off to one side or the other the angles that each tension makes with the vertical will be different and the tensions will be different. You can relate the tension in each side by equating the horizontal component of the tension on each side. that is \[ T _{l}\sin(a _{l}) = T _{r}\sin(a _{r})\] where l and r refer to the left and right side of the cable respectively .

Problems involving hanging or suspended objects via cable and struts or beams are good practice for resolving vectors.

You might try to find an mechanical engineering book on Statics which might be very helpful.