00:01
To consider my friction force, and this will be kinetic friction, will go that way.
00:10
And i've told that i've got, i'm pushing this way.
00:17
I believe this is correct at an angle of 21 degrees.
00:22
I've got m .g is my down.
00:24
My normal force is going up.
00:27
And this is my applied force right here.
00:32
This is my hand.
00:36
And my applied force goes this way.
00:40
And this will be f times the sign of 21 degrees.
00:50
And this will be f times the cosine of 21 degrees.
00:56
And we are asked to find the acceleration of the crate.
00:58
If our applied force is 3 .30 and our coefficient of friction is 0 .45.
01:07
Okay, so let's get started here.
01:09
We are going to use the following equation.
01:16
Equations, set of equations.
01:18
Let me write some down here so we can derive what we're going to be using.
01:21
F times the sign of our theta plus mg minus n has to equal zero.
01:29
So n equals f times the sign of theta plus mg.
01:37
And that's my vertical.
01:39
And then my horizontal, okay, and the horizontal will be f times the cosine of theta minus this and equals ma.
02:12
And then, so this will be f times the cosine minus this times f times the theta plus mg equals ma and a will equal f times the cosine of theta minus or coefficient of kinetic friction times f times the sign of theta plus mg divided by m.
02:56
And then we can plug our values in.
02:58
I'll have 330 times the cosine of 21 minus 0 .45 times the sign 45 times the sign, no, times 330 times the sign of 21 plus, what was my mass? i think my mass is 32 based on another problem plus 32 times 9 .81.
03:36
32 times 9 .8.
03:41
Okay.
03:44
Let me do this.
03:44
So i've got 330 330 times the cosine of 21.
03:55
I'm going to write this down in case i screw up 3 .0808 minus parentheses .45 times 330 times the sign of 21 close minus or plus 32 times 9 .81 close close.
04:28
I got an extra thing right here.
04:30
I need to go delete and hopefully this works...