Determine the kinematic equation that you would have to use.....
Vf = Vi + at
Delta x = .5(vf+vi)t
Delta x = Vi(t)+.5a(t^2)
Vf^2 = Vi^2 + 2a(delta x)
Choose which one you would use, and solve for Vf.
Consider that the Force being applied is someone pulling on a string. That string is 60.0 degrees to the flat surface the block is on. Ergo, the Force is at 60.0 degrees to the horizontal. And the friction is acting on the flat surface, keeping the angle at 180.
But, to answer your question...
You wouldn't multiply the frictional force by cos(theta) though.
If you draw your FBD....
You have a friction force going straight backwards with a magnitude of 20 N. Therefore, it has an angle of 180 with direction of motion, so cos(180) = -1. The 60.0 degrees ONLY affects the diagonal force...
You know that Velocity is a change in position over change in time, correct?
Using that, you should be able to figure out the position. (At least, that's how I do it.)
In order to come up with these equations, you have to first figure out what forces are acting upon the objects in question.
The four basic ones when dealing with pulleys are:
1. Weight (otherwise known as gravitational)
2. Normal (otherwise known as natural in some cases)
3. Tension
4...
Let's assume they are. haha
Now, add up all the equations. The tension forces will cancel leaving you with.....
\SigmaF = m_{3}g - m_{1}g = m_{1}a + m_{2}a + m_{3}a
So, factor out a g from the left side, and an a from the right side.....
g(m_{3}-m_{1}) = a(m_{1} + m_{2} + m_{3})...
Well, after you draw your FBDs, you will have a system of 3 equations.
So, sum the forces in each situation.
We will denote T_{1,2} as the tension force between the M1 and M2 object.
We will denote T_{2,3} as the tension force between the M2 and M3 object.
T_{2,1} = T_{1,2}
T_{3,2} =...
No matter how many times I do this problem, I still get .305 as uk.....
So, I don't know. Maybe there was a rounding thing that was done somewhere to make it different than .305.
I'm sorry.
I'm honestly stuck on this problem. I think I summed the forces right (since I could solve for us and get the right answer).
But, when it gets to finding the acceleration, I'm confused.
Maybe if I think about it some more.....
Alright, now we can get this problem started.
So, you'll have 3 separate FBDs.
M1's FBD will have a T force going up and a Weight Force going down.
M2's FBD will have tension forces on both sides, a normal force going up that is equal to a weight force going down.
M3's FBD will have a T...
So basically, there are two pulleys attached to the box.
There is a single, massless string run through the pulley system.
M1 is hanging on the left side of the box and M3 is hanging on the right side of the box (attached to string I presume).
M2 has a string connected on both sides? Yes?
Maybe you could multiply the 1.553 m/s^2 by cos(25.7). Since a would be a vector, and that would, technically, give you ax.
Try that and see if that works for you. I'm honestly confused.
Consider that the final velocity will be the Delta x / Delta t.
So, Vf = 3.50 m / 1.50 s
Vf = 2.333 m/s
Now, use your kinematic equation...
V_{f}=V_{i}+a*t
V_{i}=0 m/s
Now, solve for a.....
a = \frac{V_{f}}{t}
a = \frac{2.333}{1.50}
a = ?
After...
Haha. Thanks, just here to try to help.
Now, you need to find the sum of the forces in the x-direction.
Well, let's start by figuring out what N will be.
Since N is pointing straight upwards, and by Newton's 3rd law, N - Wcos(theta) = 0, right?
Now, N = Wcos(theta) -> W = mg, so...
N =...
Alright, so let me try it a different way.
Consider a normal x-y axis with N going upward directly on the y-axis and the f force going directly on the x-axis.
Now, your W force will be a diagonal line in the 4th quadrant (that is, negative y, positive x).
The angle between W and the y-axis...
Right. Now, when you draw your free body diagram, you'll notice that N is slanted in a particular direction.
You can picture the x-y axis as being slightly rotated, that way the N is now directly on the y-axis, and the f force is now directly on the x-axis.
W is now a hypotenuse going...
Then you use the acceleration and plug it into a kinematic equation.
You have delta x, acceleration, and an initial velocity.
So, you have to find your time using.....
\Deltax = V_{i}t + 1/2(a)(t^{2})
Please correct me if I'm wrong.