00:01
So we have a block of steel, but we're given its mass, and told that it is at rest on the table.
00:05
We're also given the coefficient of static friction with respect to that table.
00:10
And so first we want to calculate the magnitude of our horizontal force.
00:15
So doing a quick sketch of our free body diagram, we have our block of steel with the mass m, and we have the force of gravity acting on it, as well as a normal force.
00:28
And the force of static friction will be trying to oppose its motion.
00:36
And now what we can say is, since we only have two forces in the vertical direction, since the block is at rest, we can say that normal force is equal to mg.
00:47
Now, we can write down our equation for the force of static friction, so the force of static friction is equal to or less than our coefficient of static friction times normal force.
01:01
And so if we apply this equation, we can calculate the force needed.
01:06
And so, yeah, if we, so then we now multiply our coefficient of static friction 0 .52 times our normal force, which was mg, and we get that this is equal to our 12 kilograms times our 9 .81 gravity, and the force required will be equal to 61 .244.
01:32
So that's our first objective.
01:37
The second objective is assessing particular angles of this force.
01:46
And so if we recreate our free body diagram, we now have a force, which i'll call f, at a particular angle.
01:57
And the angle theta is above the horizontal.
02:04
And so to calculate that force required, we need to compute our sum of forces in both the x and y direction and so quickly we're going to define our positive in positive x and positive y directions we can now say that our x component will be that particular force cosine of theta minus our force of static friction and we might as well write that force of static friction as upon that formula as mu s times n the normal force and for our y direction, we'll have our normal force plus our force times sine theta minus mg is equal to zero.
03:03
And for these conditions, they're both equal to zero because we're still at rest.
03:08
And so let's go ahead and call this our system of equations.
03:14
We're going to call that first one, equation one, and the second one, equation two.
03:19
So what we're going to do now is rearrange our equation 1 to solve for our f in terms of variables.
03:31
And so if we do this, we get that f is equal to mu sub s times normal force all over cos theta.
03:42
Now what we're going to do is plug this expression for f into our equation 2.
03:49
So i'll say plug into equation two.
03:56
And if we do this, we're going to get that our normal force plus mu, mu, mu, sub s time to normal force over cosine theta times sine of theta is equal to moving the mg term to the right is equal to mg.
04:19
And so why are we doing this? well, what we're going to end up doing is solving for n.
04:32
I should say solve.
04:34
So solve for n, then solve for f.
04:44
And so the way that this will work is because if we solve for n, then we can use our equation one for the forces in the x direction to isolate f and solve for it.
04:56
Because we know theta, we know the coefficient of static friction, but we still need to compute our n.
05:05
So we can now factor this to solve for n.
05:09
And if we do this, we get n times 1 plus mu s sine theta over cosine theta is equal to mj.
05:30
So now we can rearrange this further...