F = ma Normal and Tangential Coordinates | Equations of motion| (Learn to solve any question)

Learn to solve F=ma problems with normal and tangential coordinates.

A girl having a mass of 25 kg sits at the edge of the merry-go-round...(1:15)
The 0.8-Mg car travels over the hill having the shape of a parabola...(2:50)
The block B, having a mass of 0.2 kg, is attached to the vertex A...(5:45)

Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2014.

[SG-dwjh]

My guy is the New Organic Chemistry Tutor and a little more direct and quicker! U R the Champ!!!

[camerongillespie]

I appreciate the way you walk through the math for each problem. You do a great job balancing efficiency and detail!

[joeyGalileoHotto]

By far the best videos for Dynamics on YouTube

[fahadalharbi]

In the last question, I did not understand why the tringle change between N and T. For example, you have used N (4, 3, 5), but with T (3, 4, 5), so could you explain it?

[joaobrites]

hello! i have a question regarding the second problem. For the resultant frctional force what i did was saying that the work done by all non-conservative forces must be equal to the change in the total mechanic energy of the system: Wncf= delta-Em . Because the frictional force is the only non conservative force applied we can say that it is equal to delta-Em. for the delta-Em we can call the highest point in the graph (x=0, y=20) as point "B" . lets call: kinetic_energy(point)---Ek, potential_energy(point)--Ep. thus we get : Delta-Em= Ek(a)+Ep(a)-Ek(b)-Ep(b). because the height in a=0, it follows that Ep(a)=0, and because the velocity is constant we get Ek(a)-Ek(b)=0 . Therefore : delta-Em= -Ep(b) . because of what we knew from before we say that : the work of the frictional force must be equal to -Ep(b) . we know that the work of the frictional force ( lets call it Wf ) is equal to abs(f).d.cos(alpha). we also know that the value of Ep(b) is M.H(b).g = 800x9.81x20=156960. Now, we need the value of alpha to find cos(alpha) and the value of d. alpha is always 180 degrees, because the friction force always points in the opposite direction of the movement and thus cos(alpha)=-1. now we only need D. and this is the hard part: we know that the lenght of an arc of a graph is given by the integral of (1 +(dy/dx)^2).dx - this in the interval from Xb to Xa, that is, from 0-80. We have the Y/X relationship so we only need to do the maths. I did them and i got the value of 82.72m for the arc-length ( wich is equal to d). the minus from the cos(alpha) cancels the minus from -Ep(b) and so the abs(f) is positive (as expected of course) and equal to : (156960)/82.72 = 1897, 5N .... I don´t know what i did wrong, can you help me!? thanks, great video as always!

[vikramnagarjuna]

Hi, this is vikram from India,
I like your videos. Please increase video duration by including more problems with animations.

[morolayomi]

At 5:06, could I have done 800(9.81)cos63.44 ?

[daynelissaez]

There’s an error in 5:13. The angle is 26.56 but you have used 26.57 on the equation. 😅

[muhammaddawood]

why did't you take normal axis aiming at the centre in the last question?

[morolayomi]

At 0:24 is that the approach every time? Thought normal would just be into the curve nd tangential is direction of velocity?

[swagodaman]

You channel has such an insanely good quality. thank you

[hobezie]

Thank you so much for these videos, the problems are super similar to the ones that we are going over in class and the ones in my textbook. I really appreciate it and I will def be using your other videos these are great.

[otd]

In 7:52 please can you explain which direction the normal force is acting on the block. Is it on the surface of the cone. I'm confused

[roman_surchy]

hey, my question probably would be ridiculous but I will ask anyway, could you please let me know why in the first question N is negative?

[sisyphusishappy]

In the last problem, I'm confused, why is the weight not included in the summation of force in normal axis?

[michaeltorbey]

Hello, at 7:06 why didn't you place the axes along the same direction as the cone and tension?

[chadbullard]

Can you rotate the b and n axis for last example and solve? Meaning, T and N forces would not need multiplied by angle, on weight in both force equations?

[sayanjitb]

In the last question What if I laid the the coordinate system along the slanted side of the cone?

[jbillz]

Since the block is not moving along the plane tangent to the cone, why cant we set the parallel component of gravity to the tension? It yields a different result.
Thanks,
Jacob.

[SettNaing-id]

Thank you Sir.
May Lord always bless you.