Tension Exercise on frictionless inclined plane

In summary, the problem involves two masses connected by a string and released from rest on an inclined plane. The force and free body diagrams are drawn and the equations of motion for each mass are written using Newton's Second Law. The acceleration of the masses is found to be the same and the final speed of the masses can be determined by using the tension in the string, which is found to be equal to the weight of the second mass. The only question that remains is how to find the acceleration of the second mass in the y-direction, which can be solved by using a different coordinate system for each mass.
  • #1
Kernul
211
7

Homework Statement


A mass ##m_1## is attached to a second mass ##m_2 > m_1## by an Acme (massless, unstretchable) string. ##m_1## sits on a frictionless inclined plane at an angle ##\theta## with the horizontal; ##m_2## is hanging over the high end of the plane, suspended by the taut string from an Acme (frictionless, massless) pulley. At time ##t = 0## both masses are released from rest.
a) Draw the force/free body diagram for this problem.
b) Find the acceleration of the two masses.
c) Find the tension T in the string.
d) How fast are the two blocks moving when mass m2 has fallen a height H (assuming that m1
hasn’t yet hit the pulley)?
Immagine.png


Homework Equations


Tension
Newton's Second Law

The Attempt at a Solution


First thing I drew the free body diagram this way:
photo_2016-12-19_19-15-27.jpg

Now I write the forces acting on both masses with Newton's Second Law, knowing that the accelerations of both masses are the same, so ##a = a_1 = a_2##.
First mass:
$$\begin{cases}
F_{1 x} = m_1 a_x = T - m_1 g sin \theta \\
F_{1 y} = m_1 a_y = N - m_1 g cos \theta = 0
\end{cases}$$
Second mass:
$$\begin{cases}
F_{2 x} = m_2 a_x = m_2 g sin \theta - T sin \theta \\
F_{2 y} = m_2 a_y = m_2 g cos \theta - T cos \theta
\end{cases}$$
Now what I don't understand is how I find ##a_y##. Because the only way for ##m_1 a_y = 0## is that ##a_y = 0## but this would go in contrast with the fact that in the second mass ##a_y## is something, since the mass moves along the y-axis. Or it is ##0## because the mass moves at a constant speed? In that case I would have these:
First mass:
$$\begin{cases}
a_x = \frac{T}{m_1} - g sin \theta \\
N = m_1 g cos \theta
\end{cases}$$
Second mass:
$$\begin{cases}
a_x = g sin \theta - \frac{T}{m_2} sin \theta \\
m_2 g cos \theta = T cos \theta
\end{cases}$$
With ##m_2 g cos \theta = T cos \theta## becoming ##T = m_2 g##.
Is this way correct?
 
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  • #2
Kernul said:

Homework Statement


A mass ##m_1## is attached to a second mass ##m_2 > m_1## by an Acme (massless, unstretchable) string. ##m_1## sits on a frictionless inclined plane at an angle ##\theta## with the horizontal; ##m_2## is hanging over the high end of the plane, suspended by the taut string from an Acme (frictionless, massless) pulley. At time ##t = 0## both masses are released from rest.
a) Draw the force/free body diagram for this problem.
b) Find the acceleration of the two masses.
c) Find the tension T in the string.
d) How fast are the two blocks moving when mass m2 has fallen a height H (assuming that m1
hasn’t yet hit the pulley)?
View attachment 110562

Homework Equations


Tension
Newton's Second Law

The Attempt at a Solution


First thing I drew the free body diagram this way:
View attachment 110563
Now I write the forces acting on both masses with Newton's Second Law, knowing that the accelerations of both masses are the same, so ##a = a_1 = a_2##.
First mass:
$$\begin{cases}
F_{1 x} = m_1 a_x = T - m_1 g sin \theta \\
F_{1 y} = m_1 a_y = N - m_1 g cos \theta = 0
\end{cases}$$
Second mass:
$$\begin{cases}
F_{2 x} = m_2 a_x = m_2 g sin \theta - T sin \theta \\
F_{2 y} = m_2 a_y = m_2 g cos \theta - T cos \theta
\end{cases}$$
Now what I don't understand is how I find ##a_y##. Because the only way for ##m_1 a_y = 0## is that ##a_y = 0## but this would go in contrast with the fact that in the second mass ##a_y## is something, since the mass moves along the y-axis. Or it is ##0## because the mass moves at a constant speed? In that case I would have these:
First mass:
$$\begin{cases}
a_x = \frac{T}{m_1} - g sin \theta \\
N = m_1 g cos \theta
\end{cases}$$
Second mass:
$$\begin{cases}
a_x = g sin \theta - \frac{T}{m_2} sin \theta \\
m_2 g cos \theta = T cos \theta
\end{cases}$$
With ##m_2 g cos \theta = T cos \theta## becoming ##T = m_2 g##.
Is this way correct?
I can't really see the details of your diagram too well, but why are you assuming that block 2 has an acceleration in the x-direction?
 
  • #3
Kernul said:

Homework Statement


A mass ##m_1## is attached to a second mass ##m_2 > m_1## by an Acme (massless, unstretchable) string. ##m_1## sits on a frictionless inclined plane at an angle ##\theta## with the horizontal; ##m_2## is hanging over the high end of the plane, suspended by the taut string from an Acme (frictionless, massless) pulley. At time ##t = 0## both masses are released from rest.
a) Draw the force/free body diagram for this problem.
b) Find the acceleration of the two masses.
c) Find the tension T in the string.
d) How fast are the two blocks moving when mass m2 has fallen a height H (assuming that m1
hasn’t yet hit the pulley)?
View attachment 110562

Homework Equations


Tension
Newton's Second Law

The Attempt at a Solution


First thing I drew the free body diagram this way:
View attachment 110563
Now I write the forces acting on both masses with Newton's Second Law, knowing that the accelerations of both masses are the same, so ##a = a_1 = a_2##.
First mass:
$$\begin{cases}
F_{1 x} = m_1 a_x = T - m_1 g sin \theta \\
F_{1 y} = m_1 a_y = N - m_1 g cos \theta = 0
\end{cases}$$
Second mass:
$$\begin{cases}
F_{2 x} = m_2 a_x = m_2 g sin \theta - T sin \theta \\
F_{2 y} = m_2 a_y = m_2 g cos \theta - T cos \theta
\end{cases}$$
Now what I don't understand is how I find ##a_y##. Because the only way for ##m_1 a_y = 0## is that ##a_y = 0## but this would go in contrast with the fact that in the second mass ##a_y## is something, since the mass moves along the y-axis. Or it is ##0## because the mass moves at a constant speed? In that case I would have these:
First mass:
$$\begin{cases}
a_x = \frac{T}{m_1} - g sin \theta \\
N = m_1 g cos \theta
\end{cases}$$
Second mass:
$$\begin{cases}
a_x = g sin \theta - \frac{T}{m_2} sin \theta \\
m_2 g cos \theta = T cos \theta
\end{cases}$$
With ##m_2 g cos \theta = T cos \theta## becoming ##T = m_2 g##.
Is this way correct?
you do not have to use the same coordinate system for the two masses. You may use a vertical y-axis for mass 2 in which case there are only forces along y for mass 2 and the equation is simple. The key point is to use indices to distinguish the accelerations of the two masses, you should use ##a_{x1}, a_{y1}, a_{x2},a_{y2}##. You should notice that then the values of ##a_{x1}## and ##a_{y2}## are not independent, they are related by a simple equation.
 
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  • #4
Elvis 123456789 said:
I can't really see the details of your diagram too well, but why are you assuming that block 2 has an acceleration in the x-direction?
It's because I used the same diagram for both masses, so I end up having the second mass moving on the x-axis too.

nrqed said:
you do not have to use the same coordinate system for the two masses. You may use a vertical y-axis for mass 2 in which case there are only forces along y for mass 2 and the equation is simple. The key point is to use indices to distinguish the accelerations of the two masses, you should use ##a_{x1}, a_{y1}, a_{x2},a_{y2}##. You should notice that then the values of ##a_{x1}## and ##a_{y2}## are not independent, they are related by a simple equation.
But I've been taught that since the two masses are linked by a string, we should use one single coordinate system for both.
Anyway, following your way, the first one remains the same and the second one becomes like this:
$$\begin{cases}
a_{x 2} = 0 \\
a_{y 2} = g - \frac{T}{m_2}
\end{cases}$$
Being not independent because of the string, ##a_{x 1}## and ##a_{y 2}## are equal and so we have:
$$\frac{T}{m_1} - g sin \theta = g - \frac{T}{m_2}$$
With some passages I end up with:
$$T = \frac{m_1 m_2}{m_1 + m_2} g (1 + sin \theta)$$
Substituting this into one of the two I get the acceleration and substituting it into this equation ##v_f = \sqrt{2 a H}## I get the last point of the exercise.
Am I right?
 
  • #5
Kernul said:
It's because I used the same diagram for both masses, so I end up having the second mass moving on the x-axis too.But I've been taught that since the two masses are linked by a string, we should use one single coordinate system for both.
Anyway, following your way, the first one remains the same and the second one becomes like this:
$$\begin{cases}
a_{x 2} = 0 \\
a_{y 2} = g - \frac{T}{m_2}
\end{cases}$$
Being not independent because of the string, ##a_{x 1}## and ##a_{y 2}## are equal and so we have:
$$\frac{T}{m_1} - g sin \theta = g - \frac{T}{m_2}$$
With some passages I end up with:
$$T = \frac{m_1 m_2}{m_1 + m_2} g (1 + sin \theta)$$
Substituting this into one of the two I get the acceleration and substituting it into this equation ##v_f = \sqrt{2 a H}## I get the last point of the exercise.
Am I right?
Looks good!
 
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FAQ: Tension Exercise on frictionless inclined plane

What is tension exercise on a frictionless inclined plane?

Tension exercise on a frictionless inclined plane is a physics experiment that involves studying the relationship between tension, weight, and angle on an inclined plane with no friction. It is a common exercise in introductory physics courses to demonstrate concepts such as Newton's laws of motion and the use of free-body diagrams.

How does friction impact tension exercise on a frictionless inclined plane?

Friction does not have an impact on tension exercise on a frictionless inclined plane because it is assumed to be absent. In this type of exercise, the inclined plane is considered to be perfectly smooth and without any frictional forces acting on it.

What are the key variables in tension exercise on a frictionless inclined plane?

The key variables in tension exercise on a frictionless inclined plane are the weight of the object being pulled, the angle of the incline, and the tension force acting on the object. These variables are used to calculate the acceleration of the object down the inclined plane.

How is tension calculated in this type of exercise?

Tension can be calculated using the formula T = mg sinθ, where T is the tension force, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the incline. This formula takes into account the weight of the object and the angle of the incline.

What are the practical applications of tension exercise on a frictionless inclined plane?

Tension exercise on a frictionless inclined plane has practical applications in fields such as engineering, where it can be used to calculate the forces acting on objects on inclined planes. It can also help in understanding the mechanics of objects on inclined surfaces, such as ramps or hills, and can be used to optimize the design of structures such as roller coasters or ski slopes.

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