Recent content by sb13

Oh. I did multiply by gravity twice. Ok. Thank you!

So, I understand that I will just be using variables for the mass, but I don't understand how I use the fk and fs to find the acceleration. Once I get FN= ((9.8)mg)- (0.5(mgsin20)) and then times it by the coefficient of fs then what am I supposed to do with the value of fs?

Homework Statement There is an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle θ = 20°. What is the magnitude of the acceleration of the block across the floor if (a)μs = 0.630 and μk = 0.540 and (b)μs = 0.400 and μk = 0.330...

Homework Statement Suppose that in the figure shown in the URL below, the masses of the blocks are M = 2.6 kg and m = 6.6 kg. What are (a) the magnitude of the acceleration and (b) the tension in the...

Ok. So the gravitational force ends up creating a greater force and because the acceleration is going upward, it is causing more force as well? What is the acceleration was -2.0 m/s^2 and the plane was being pulled downward, what effect would that have on the engine's force?

Homework Statement A 1600 kg jet engine is fastened to the fuselage of a passenger jet by just three bolts (this is the usual practice). Assume that each bolt supports one-third of the load. (a) Calculate the force on each bolt as the plane waits in line for clearance to take off. (b) During...

Nope. I finally understand it completely.

Oh. So I was not supposed to square the (3600 * 24) then I guess. Thanks.

Ok. So does that mean that the equation that I choose to use to determine the velocity should include the distance (x)?

Well, I did end up getting the correct velocity. But I am wondering why I couldn't get the right answer with the v=vo + at formula v=vo + at= (0.029487m/s^2)(1s)? Is one second the time? Then the velocity equals the acceleration, and that's not the right answer. I tried v^2= vo^2 +...

The only difference is that I carried out the significant digits on my acceleration. (0.5)(0.029487)(3600*24)^2 = 1.1006 X 10^8

You didn't get exactly 1.1006X10^8 for the distance? But, I do need the velocity to be in m/s.

I got the correct distance of 1.1006 X 10^8, and for the time I converted seconds squared to days squared by multiplying by (3600^2)(24^2). In the equation of x-xo = volt+ 1/2at^2, I assumed that the initial velocity and initial distance was zero. Is that correct?

So for (c) I tried the equation v=vo + at to find the velocity, but it is not giving the correct answer. The initial velocity is equal to zero, and the acceleration is equal to 0.029487 m/s^2, but I want the time to be in seconds so that it will cancel off one of the seconds leaving just m/s...

Oh! Thank you so much. That was my main problem. I just was worried because I ended up with days^2 and I thought that would mess things up.