Friction, Inclined Plane, and Pulley HW question?
Follow
 ✓Follow publicly
 ✓Follow privately
 Unfollow
Q: You have an inclined plane with a pulley on top. Two masses are coming off the pulley. The 10kg mass is on the slope. The 50kg mass is hanging off the side of the inclined plane. ...show more
Update : And I don't know if this effects anything, but we never learned ...show more
Best Answer
These problems are modified versions of the Atwood machine.
They are best approached looking at free body diagrams (FBDs) of each component and then applying F=m*a
When the pulley is massless and frictionless, there are some simplifications to the problem.
First, the tension in the cord is the same everywhere
(there is a special case where that is not true. I won't go into it here since it may confuse. If you want to know more about the case of "slack cord", send me an email).
Second, the acceleration of the masses are equal.
Okay, I like to start with a FBD of the hanging mass. Since it is much more massive than the one on the slope, I will make the assumption that the mass will descend when released from rest if the tension is sufficient to break static friction on the slope. So positive will be in that direction of motion
50*9.81T=50*a
T=50(9.81a)
Now let's look at the mass on the slope.
The forces parallel to the slope are
T, the tension in the cord
10*9.81*sin(30), the force of gravity parallel to the slope
10*9.81*cos(30)*µ, this is the force of friction
now let's set up F=m*a and test to see if the system will accelerate.
T10*9.81*sin(30)
10*9.81*cos(30)*µ=10*a
simplify
T=10*a+49.05+84.96*µ
from above
50*(9.81a)=10*a+49.05+84.96*µ
solve for a
a=(441.4584.96*µ)/60
Ok,
let's test to see if we get a>0 when µ=0.6
a=6.51 so it will move
Once moving, µ=0.5, so the actual
a=6.65
Just for fun, let's find T from the FBD of the hanging mass
158 N
Check by plugging T and a into the FBD equation for
the mass on the incline
You will get the two sides equal if all of the math and algebra are done right. I get 66.5=66.5
j
They are best approached looking at free body diagrams (FBDs) of each component and then applying F=m*a
When the pulley is massless and frictionless, there are some simplifications to the problem.
First, the tension in the cord is the same everywhere
(there is a special case where that is not true. I won't go into it here since it may confuse. If you want to know more about the case of "slack cord", send me an email).
Second, the acceleration of the masses are equal.
Okay, I like to start with a FBD of the hanging mass. Since it is much more massive than the one on the slope, I will make the assumption that the mass will descend when released from rest if the tension is sufficient to break static friction on the slope. So positive will be in that direction of motion
50*9.81T=50*a
T=50(9.81a)
Now let's look at the mass on the slope.
The forces parallel to the slope are
T, the tension in the cord
10*9.81*sin(30), the force of gravity parallel to the slope
10*9.81*cos(30)*µ, this is the force of friction
now let's set up F=m*a and test to see if the system will accelerate.
T10*9.81*sin(30)
10*9.81*cos(30)*µ=10*a
simplify
T=10*a+49.05+84.96*µ
from above
50*(9.81a)=10*a+49.05+84.96*µ
solve for a
a=(441.4584.96*µ)/60
Ok,
let's test to see if we get a>0 when µ=0.6
a=6.51 so it will move
Once moving, µ=0.5, so the actual
a=6.65
Just for fun, let's find T from the FBD of the hanging mass
158 N
Check by plugging T and a into the FBD equation for
the mass on the incline
You will get the two sides equal if all of the math and algebra are done right. I get 66.5=66.5
j
Other Answers (1)

it is through practise you will be able to solve such problems.
First you have to find out whether the mass on the inclined plane will move or not. It will move only if the tension in the string due to hanging mass exceeds the limiting static frictional force of the mass on the plane.
Assuming that all the masses are stationary, the tension in the string is equal to weight of the handing mass = 490N.
Maximum static frictional force for the mass on the plane is
.6*Mg*cos 30= 50.9N.
As 490N>50.9N, we can conclude that the masses will move. Due to motion the frictional force on the mass on the slope will be less and it will be .5*Mg*cos 30= 42.48.
Now you have to draw free body diagrams for both the masses and eliminate the uknown Tension in the string and find the acceleration of the hanging mass. As it is your homework i will let you do that :)Source(s):
no particular source
Sign In
to add your answer