00:01
In this problem, we have a free -rolling ramp with a given mass and a crate that is placed at the top of the ramp at point a and slide down to point b.
00:11
If the surface of the ramp is smooth, we want to determine the ramp speed when the crate reaches b and also the velocity of the crate at the crater point.
00:19
So firstly, we'll apply the conservation of energy.
00:29
Now we know that the kinetic energy of the system, the ramp and the crate at point a, t1, plus its gravitation, potential energy v1 must equal to the kinetic energy plus potential energy at point b when the crate reaches point b t2 plus v2.
00:50
Now we'll set the datum at the lowest point b so when the crate is at a we know it's gravitational potential energy and its kinetic energy is zero since the crate and ramp about that rest the gravitational potential energy is 3 .5 times the sign 30 degrees times g 9 .81 times the mass 10 kges.
01:33
And this potential energy, this gravitational potential energy, then converted to kinetic energy at b, that's the kinetic energy of the crate that's a half times the mass.
01:46
Times the speed of the crate squared plus the kinetic energy of the ramp, which is a half, and its mass 40 kg times its speed bb squared.
02:00
And so we end up with 171 .675 is equal to 5 times bf squared plus 20 bb squared.
02:18
So we have an equation with two unknowns, the speed of the crate and the speed of the ramp.
02:29
Vb will use the r for the speed of the ramp.
02:34
So we have an equation with two unknowns.
02:38
Now we'll apply relative velocities.
02:48
So applying relative velocities, we get the velocity of the crate.
02:55
Vc, the vector velocity is equal to the velocity of the ramp, vr plus the velocity of the crate relative to the ramp vcr and so this is equal to minus vr in the i direction plus vc relative to r times the cosine of 30 degrees now we break it into its i and j components that i minus the magnitude vc relative to r times the sign of 30 degrees and that's in the j direction or the y axis so this simplifies to 0 .8 660 times the magnitude of the cr minus vr in the i direction minus 0 .5 in the j direction and we'll call this equation 2.
04:17
Now we have two unknowns in that equation as well.
04:23
Now from here we can actually find the magnitude vc.
04:28
So the magnitude vc is simply the square root of these components we calculated above.
04:39
So that's and the square root of the components squared and summed.
04:46
So 0 .866 v .r minus vr squared plus minus 0 .5 v .r squared...