Physics help please
I feel a little lost. Here's the scenario. There are two objects m and M are connected by a string that goes through a pulley on an incline.
The incline is 45 degrees from the horizontal. m = 4Kg and M = 10Kg and theta (@) = 45 degrees. I am assuming that m is only the inclined plane and M is hanging on the string. Do not consider friction.
a) what forces are acting on each object?
I got the horizontal and vertical components of the gravitational force on m. I got Fy = mgsin@ and Fx = mgcos@.
For m, I have F in the y = Normal force - mg(sin@). F in the x = Tension - ma(cos@)
For M, I have F = T - Mg.
b) calculate the accel. for the objects
For M, I wrote that the accel for M is gravity (-9.8m/ss). If the system is still at rest, T = Mg.
For m, I supposed that m had a net force of 0 on the y direction because the object isn't sinking into the plane nor it is flying up.
For x direction, I had ma = T - mg(cos@). I solved for a and got a = (Mg - mg(cos@))/m
Is this correct?
c) calculate tension
Since I had T = Mg from the previous step, I just plugged in the numbers and that's that.
d) how fast are the objects after 5sec from rest.
I have t = 5s, initial v = 0m/s, and the a = the answer from part b.
I used final v = initial v - at.
Can you guys confirm if I am on the right track?
5 Comments • Newest first
a):
You mixed up your sin@ and cos@ for your vertical and horizontal components of gravity. In this case since the angle was 45 degrees it didn't matter, but for future reference the Y component would be m*g*cos@ and the X is m*g*sin@. It helps a lot to draw a free body diagram when identifying forces and their components. Aside from that, I'm sure you meant to say "mg" instead of "ma" in Tension - macos@. It would correctly be written as mgsin@. Your forces for M are correct.
b):
I believe you misunderstood how pulleys work. When two masses are connected by a pulley, when one mass moves, the other moves with the same magnitude. Therefore when either M or m move with acceleration [i]a[/i] , the other would move with the same magnitude of acceleration. One can think of the two masses as a single system. For these kinds of questions, they are trying to get you to find:
1. In which direction will the system move (will m fall down and M up or M fall down and m up)?
2. What is the magnitude of the acceleration?
A really easy way to think of simple pulley systems like this is to combine the masses and look at it as if it were a single object. Pulling on one side is the force of gravity on mass M. Pulling on the opposite side is the component of the force of gravity on mass m. The forces of tension are internal forces, so they can be ignored. After that you can just use Fnet = m*a to find the magnitude of acceleration, and whichever mass had a greater force of gravity (or component of gravity in the case of mass m) is the direction the system accelerates.
c):
You cannot use T=mg anymore because we know that the masses must be accelerating in a certain direction, meaning the net force cannot be zero therefore T cannot = mg. Just use the magnitude of acceleration you just found and plug it into Fnet=m*a for a single mass (you can choose between m and M). If we use M as an example, the force of gravity is Mg and net force is Ma. Since Fnet = Ftension + Fgravity (assume I'm using vectors here), Ftension = Fnet - Fgravity. This can now easily be solved.
d):
Kinematics is easy as hell. You don't need my help here.
If you have any more questions, feel free to ask.
yes, I kinda suspected it too
I forgot to add. I was also assuming block m is suspended in the air by the rope.
[quote=HolyDragon]Sorry I can't be of much help.
I'll throw a guess.
Assuming it's ordered like
m---M
Slope is uphill in the positive x direction.
A
Isn't the horizontal force for the incline plane sin, not cos? Likewise with vertical force. Either way, since it's a 45 degree triangle, their values are the same.
Also, I'm assuming the slope is 45 degrees from a first quadrant point of view. So isn't it Tension + the force doing downhill? With no friction, they shouldn't be a force uphill.
If we don't consider friction, I don't think this is at rest.
B. I would actually guess it as one system
(m+M)a= (Mg + msin(45))
This acceleration would be to the left.
C. Edit: Actually, since it's getting pulled and pushed, I think it's MG - msin(45)
D. velocity and acceleration formula.[/quote]
Oh crap lol I fudged up with the trig! Thanks, I knew something felt off!
A & B: Yes, it's all in the 1st quadrant PoV. Since the heavier object is pulling on the lighter object on the incline, I supposed that the Tension and the horizontal component of the gravitational force would act against each other. Basically, mg(sin@) is going negative x and T is going positive x with the incline being the x axis. Thanks again...I calculated this as though they're not part of the same system.
Man, I feel dumb...
Sorry I can't be of much help.
I'll throw a guess.
Assuming it's ordered like
m---M
Slope is uphill in the positive x direction.
A
Isn't the horizontal force for the incline plane sin, not cos? Likewise with vertical force. Either way, since it's a 45 degree triangle, their values are the same.
Also, I'm assuming the slope is 45 degrees from a first quadrant point of view. So isn't it Tension + the force doing downhill? With no friction, they shouldn't be a force uphill.
If we don't consider friction, I don't think this is at rest.
B. I would actually guess it as one system
(m+M)a= (Mg + mgsin(45))
This acceleration would be to the left.
C. Edit: Actually, since it's getting pulled and pushed, I think it's MG - mgsin(45)
D. velocity and acceleration formula.