00:01
Here we are given that both boxes move together at a constant speed of 15 cm per second so if we clearly examine this picture there is a plank and there are two boxes and the boxes are being pulled by person by applying force f so the below boxes of 48 cages and the above box is of 32 pages and the the kinetic friction between the below box and the plank is given to be 0 .444.
00:41
Let's say mu k equal to 0 .444 and the static friction between upper box and the lower box mu s is given to be 0 .8.
00:56
So here we are asked two questions.
00:59
First question is what force do we need to exit to accomplish this? so, in other words, they are asking what is the required f to produce the 15 cm per second and this kinetic friction of 0 .4 and static friction of 0 .8.
01:19
So, and the second question they are asking is what are the magnitude and direction of the force on the upper box.
01:28
So let's start with the first question here.
01:32
So by using the trigonometry, we can find the angle of the...
01:38
Plank so that is let's call angle theta equal to tan inverse 2 .5 by 4 .75 that will give us similar to 27 .76 degrees so then we will be calculating the normal force so if we clearly see this will be the mg and if we decompose there will be two components of this this is the normal component so the normal component will be m g cost theta so n equal to m g cost theta as m equal to 32 plus 48 that will be 80 kilograms if you substitute in this 80 times is 9 .8 that is acceleration due to gravity and cost 27 .76 degrees this will be give us a similar result as 694 .5 newton and then if we see the freeboardy diagram and write the equations of force the equations of force will be the force we are applying plus the force due to friction will be equal to m g sine teta as the force and the friction force in this direction and the force due to gravity will be in this direction that will be mg sin theta so f will be equal to mg sin theta minus mu and that will be equal to m g sine theta minus mu n mg cost theta as we have seen that normal force is equal to mg cost theta mg sin theta mg xan theta minus the kinetic friction coefficient of kinetic friction so it will be equal to 80 times 9 .8 times sine 27 .76 minus as mu qu is 0 .444 and cost of 27 .76 degrees.
05:31
So this will give us the result similar to 57 .2 newtons.
05:41
So the answer for question will be, first question is 57 .2 neutrons.
05:48
So for next question b, we need to find the magnitude of force being applied on upper plank and the direction of the fixture force.
06:00
So as we are talking about the upper box being stationary, like it has a static friction coefficient, so upper box doesn't accelerate.
06:19
So it is in static motion.
06:26
So let's call it fr and it will be equal to m .g...