00:01
In a warehouse, a wagon with a mass of 250 kilograms must be transported over a ramp.
00:09
The height of the ramp is 1 .5 meters and the slope angle is 5 degrees.
00:14
The rolling friction coefficient of the wheels is 0 .5.
00:21
We can neglect the moment of inertia of the wheel.
00:25
Question 8, you want to calculate the frictional force which acts on the wagon during rolling on the ramp.
00:34
Let's draw illustration.
00:36
I'm going to exaggerate the angle of the wheel.
00:38
Ramp so our ramp at the end is at a height of 1 .5 meters and here we have an angle of 5 degrees so when our wagon is being pushed up the ramp it is being weighted down by gravity and what's opposing the upwards motion is a downwards force of friction and its force of friction is related as usual to the normal force let's place our corner system and client and let's sum the forces along y in order to find the normal force.
01:41
So the sums of the forces along y must be equal to zero, which means that our normal force is equal to m g coast theta.
01:55
And from here we can calculate our friction force, which is given by our coefficient of friction times the normal force.
02:22
So we obtain mu and g cost data.
02:28
And just to elaborate on the variables here, um, i'll label the height as 1 .5 meters the angle ramp is 5 degrees and hence the length of our ramp can be found as untrigger will have the pythagorean theorem the hypotenuse will be given by 1 .5 over sign data which is 17 .21 meters and next we want to find or let's label we have the mass which is 250 kilograms and the coefficient of friction which is 0 .0.
03:38
0 .05.
03:41
The rolling friction, so kinetic friction.
03:45
And putting all these variables, we obtain a friction force of 122, putting 0 .03.
04:00
Mows assume that the wagon is at the top of the ramp and starts to roll down from rest.
04:05
We want to determine the velocity at the end of the ramp.
04:09
So here we're going to look at our kinematic expressions...