# Triangular Trolley: Solving Mass & Force Dynamics

• Nexus99
In summary: A string wrapped around a pulley will exert a force on the pulley, the magnitude and direction of which can be determined by considering two tension forces on the sub-system consisting of the pulley and the section of string in contact with the pulley.
Nexus99
Homework Statement
A trolley with a triangular section, as in figure (angles at the base α and β), has a
mass M and rests on a horizontal plane without friction, on which it is free of
move. On the inclined planes corresponding to two of its sides are supported two
masses m1 and m2. These are connected together by an inextensible and massless wire, and
they can slide freely and without friction. Determine the acceleration of the trolley. Especially consider the case α = β.
Relevant Equations
Newton's principles

Tried to solve it in this way:For mass M i used a reference with x-axis parallel to the ground:
## Ma = N_1 sin α − N_2 sin β ##

While for mass ## m_1 ## and ##m_2## i choose two different references with x-axis parallel to the plane and y-axis perpendicular to it.

Force perpendicular to the plane:
Relative acceleration is zero, so acceleration is equal to the projection of the absolute acceleration
## m_1 (−a sin α) = N_1 − m_1g cos α ##
## m_2 (a sin β) = N_2 − m_2g cos β ##

Force parallel to the plane:
In this case we have a relative acceleration
## m_1(a cos α + a_x) = T - m_1 g sin α ##
## m_2(a cos β + a_x) = - T + m_2 g sin β ##

Is it right for now, or am i missing something?

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Nexus99 said:
Is it right for now
Looks good so far.

etotheipi
There is no need to write so many equations with so many reaction forces.
The center of mass theorem in horizontal direction+energy conservation +kinematics equation are enough for a two degrees of freedom system

etotheipi
wrobel said:
There is no need to write so many equations with so many reaction forces.
The center of mass theorem in horizontal direction+energy conservation +kinematics equation are enough for a two degrees of freedom system
It will be necessary to subdivide at some point since the question asks for the acceleration of the triangular trolley.

It will be necessary perhaps to differentiate the energy integral. I do not know whether it is a problem for OP

By solving that system:

## Ma = N_1 sin α − N_2 sin β ##
## m_1 (−a sin α) = N_1 − m_1g cos α ##
## m_2 (a sin β) = N_2 − m_2g cos β ##

i got:
## N_1 = m_1 (g cos α - a sin α) ##
## N_2 = m_2 (a sin β + g cos β) ##
## a = \frac{m_1 cos α sin α - m_2 cos β sin β}{M + m_1sin^2 α + m_2 sin^2 β} g ##

But the result should be:
##a = \frac{(m_1 cos α + m_2 cos β)(m_1 sin α − m_2 sin β)}{M(m_1 + m_2) + m_1m_2(cos α − cos β)^2 + (m_1 + m_2) (m_1 sin^2 α + m_2 sin^2 β)} g ##

wrobel said:
It will be necessary perhaps to differentiate the energy integral. I do not know whether it is a problem for OP
Never tried to differentiate the energy integral, but looks feasible

@wrobel is certainly correct that it is simpler to write the equation for the total horizontal acceleration of the system than to bother with the normal forces. Combining that with your other equations I get the book answer, so I suspect your error is in algebra you have not posted.

haruspex said:
@wrobel is certainly correct that it is simpler to write the equation for the total horizontal acceleration of the system than to bother with the normal forces. Combining that with your other equations I get the book answer, so I suspect your error is in algebra you have not posted.
I don't know where is the error:

## Ma = N_1 sin α − N_2 sin β ##
## m_1 (−a sin α) = N_1 − m_1g cos α ##
## m_2 (a sin β) = N_2 − m_2g cos β ##

i got:
## N_1 = m_1 (g cos α - a sin α) ##
## N_2 = m_2 (a sin β + g cos β) ##
## Ma = m_1sin α(gcos α - a sinα) - m_2sin β( g cosβ + a sin β)##

## a(M + m_1sin^2 α + m_2 sin^2 β) = m_1sin α cos α g - m_2 sin βcosβ g ##

## a = \frac{m_1sin α cos α - m_2 sin βcosβ}{M + m_1sin^2 α + m_2 sin^2 β} g ##

Your equation for Ma treats the normal forces as though they are horizontal.

Edit: cancel that... see post #12.

Last edited:
haruspex said:
Your equation for Ma treats the normal forces as though they are horizontal.

Nexus99 said:
Sorry, must be going blind.
No, the error is that you have forgotten about the forces exerted on the wedge by the string passing over its peak.

etotheipi
Can the
haruspex said:
Sorry, must be going blind.
No, the error is that you have forgotten about the forces exerted on the wedge by the string passing over its peak.
Can the string exert a force in this way? Which is verse and direction?

Nexus99 said:
Can the string exert a force in this way? Which is verse and direction?

A string wrapped around a pulley will exert a force on the pulley, the magnitude and direction of which can be determined by considering two tension forces on the sub-system consisting of the pulley and the section of string in contact with the pulley.

That said, probably the easiest way to get the relation you need is to transform into the wedge frame and equate the rate of change of momentum of the entire system to the total fictitious force (N.B. here I have used ##\hat{x}'## aligned with the lab frame)$$F'_{ext, x'} = \frac{d}{dt} P'_{x'}$$ $$-(m_1 + m_2 + M)a = \frac{d}{dt} (m_1 v'_{1,x'} + m_2 v'_{2,x'})$$and you can use that in this frame ##||\vec{a}_1'|| = ||\vec{a}_2'||## to find that, using your definition of ##a_x##, the components ##a'_{1,x'} = a_x \cos{\alpha}## and ##a'_{2,x'} = a_x \cos{\beta}##

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Nexus99
etotheipi said:
A string wrapped around a pulley will exert a force on the pulley, the magnitude and direction of which can be determined by considering two tension forces on the sub-system consisting of the pulley and the section of string in contact with the pulley.

That said, probably the easiest way to get the relation you need is to transform into the wedge frame and equate the rate of change of momentum of the entire system to the total fictitious force (N.B. here I have used ##\hat{x}'## aligned with the lab frame)$$F'_{ext, x'} = \frac{d}{dt} P'_{x'}$$ $$-(m_1 + m_2 + M)a = \frac{d}{dt} (m_1 v'_{1,x'} + m_2 v'_{2,x'})$$and you can use that in this frame ##||\vec{a}_1'|| = ||\vec{a}_2'||## to find that, using your definition of ##a_x##, the components ##a'_{1,x'} = a_x \cos{\alpha}## and ##a'_{2,x'} = a_x \cos{\beta}##
Thanks, i'll try to do the problem in that way.
In the meantime i modified the first equation:

##Ma = N_1 sin α − N_2 sin β − T cos α + T cos β##
and i got the result from the book, in particular, when ## \alpha = \beta ##

## a = \frac{(m_1 − m_2) sin α cos α}{M + (m1 + m2) sin^2α}g ##

haruspex said:
Your equation for Ma treats the normal forces as though they are horizontal.
Nexus99 said:
Thanks, i'll try to do the problem in that way.
In the meantime i modified the first equation:

##Ma = N_1 sin α − N_2 sin β − T cos α + T cos β##
and i got the result from the book, in particular, when ## \alpha = \beta ##

## a = \frac{(m_1 − m_2) sin α cos α}{M + (m1 + m2) sin^2α}g ##
yes, that correctly accounts for the forces exerted on the wedge.
But, pace @etotheipi , you don't need to switch to the wedge frame to take the whole of system view. Equivalently, just use the relative accelerations.
##(M+m_1+m_2)a+m_1a_x\cos(\alpha)+m_2a_x\cos(\beta)=0##.
This replaces all equations involving forces between the masses/string system and the wedge.

etotheipi

## What is the Triangular Trolley and how does it work?

The Triangular Trolley is a simple machine used to study mass and force dynamics. It consists of a triangular-shaped trolley with wheels and a pulley system. The trolley is attached to a string that runs over the pulley and is connected to a hanging mass. When the hanging mass is released, it pulls the trolley forward, allowing for the study of mass and force dynamics.

## What is the purpose of using a Triangular Trolley?

The Triangular Trolley is used to demonstrate the relationship between mass, force, and acceleration. By changing the mass of the trolley or the hanging mass, students can observe how these variables affect the acceleration of the trolley.

## How is the acceleration of the trolley calculated?

The acceleration of the trolley can be calculated using the formula a=F/m, where a is the acceleration, F is the force applied, and m is the mass of the trolley. By measuring the force applied and the mass of the trolley, students can calculate the acceleration and observe how it changes with different variables.

## What are some real-world applications of understanding mass and force dynamics?

Understanding mass and force dynamics is crucial in many fields, including engineering, physics, and mechanics. It is used to design and improve machines, vehicles, and structures, as well as to understand the movement of objects in nature, such as the motion of planets and satellites.

## Are there any limitations to using the Triangular Trolley for studying mass and force dynamics?

While the Triangular Trolley is a useful tool for understanding mass and force dynamics, it does have some limitations. It assumes that all forces acting on the trolley are parallel, which may not always be the case in real-world scenarios. Additionally, friction and air resistance are not considered, which may affect the accuracy of the results.

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