Recent content by Pascal1p

Homework Statement Homework Equations F=m*a ΣF(n-direction)= m*a(n)= m*(v^2/ρ) ∑F(t-direction)= m*a(t)\ Ffriction= Fn*coefficient(of friction) The Attempt at a Solution So I tried solving this question and apperently it is way easier than I thought. So I thought the kinetic friction has a...

Homework Statement Homework Equations f=m*a V= Vg*cos θ The Attempt at a Solution Is this solution correct. I did everything the same, but I did not use the 15 ft/s. Since An= (v^2/ρ) with V being the velocity in the t direction. The 15 ft/s according to me is only the component of V that...

That minimum value of d, does it correspond with the answer they give? I don't get that they do Fspring= k*(Yeq + Y) and then say if Y=d then N=0. Because like you said N=0 is at unstretched position, so when Fspring= 0 Fspring = k*(Yeq+d) that corresponds to the springforce on the pan, when the...

Homework Statement Homework Equations F=m*a The Attempt at a Solution Don't worry it not a homework question of mine. I was just trying to solve some questions (on my own) that I know I have difficulty with to get better at it. Especially with springs I can get confused, so I thought I...

But even the solution model says that the value of this m*a equals 2*9.81*sin(45). They use: m*a(along shaft)= 2*9.81*sin45 = 2(2cos45 + a[c relative to ab]). Even if it not stuck to the shaft, the only force that can make it slide along shaft is the gravitational force, which won't change...

Homework Statement The 2-kg collar C is free to slide along the smooth shaft AB. Determine the acceleration of collar C if (a) the shaft is fixed from moving. (b) Collar A, which is fixed to shaft AB, moves downward at constant velocity along the vertical rod, and (c) collar A is subjected to a...

So 0.5*5*(0.2)^2+(0.203-0.2)^2*5= I(cylinder) Together with the I(plate) gives 0.227. So why is it still off?

Ow, so the moment of inertia for the semi cylinder with (1/2)MR^2 is not around the the center of the semi cylinder but around the center of the full cylinder it would make? So in this case it would be around the z-axis which passes through y=200 mm?

Homework Statement Determine the location of y of the center of mass G of the assembly and then calculate the moment of inertia about an axis perpendicular to the page and passing through G. The block has a mass of 3 kg and the mass of the semicylinder is 5 kg. Homework Equations -parallel...

No one :(?

So how to simplify this further to r^4/8(a-sin(a))

This is to where I have simplified it:

Yeah you are right, I see now how the integral with sin^-1 and tan^-1 are the same. But so that whole integral 2∫(y^2√(R^2-y^2)+y^3/tan(a/2))dy with limits 0 to Rsin(a/2) will clean up to r^4/8(a-sin(a))? Because that I do not see.

Well, the link you send, in which you said it did not do that for you. It also gives an arcsin and looked similair to what book give so I thought they were same without really checking. I did miss a parenthesis in the end, but still what the book says does not seem the same.

But where then does this inconsistency come from? My book gives some standard integrals and one of them was ∫x^2√(a^2-x^2)dx= -(x/4)√((a^2-x^2)^3)+ (a^2/8)(x√(a^2-x^2)+a^2sin^-1 (x/a) + C, a>0 The weird thing is that wolfram-alpha gives this long answer when you substitute 'a' with an actual...