Triangular Trolley: Solving Mass & Force Dynamics

[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?

 

[haruspex]

Is it right for now

Looks good so far.

 

[wrobel]

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

 

[haruspex]

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.

 

[wrobel]

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

 

[Nexus99]

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:
$a = \frac{(m_1 cos α + m_2 cos β)(m_1 sin α − m_2 sin β)}{M(m_1 + m_2) + m_1m_2(cos α − cos β)^2 + (m_1 + m_2) (m_1 sin^2 α + m_2 sin^2 β)} g$

 

[Nexus99]

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

 

[haruspex]

@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.

 

[Nexus99]

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

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

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

 

[haruspex]

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

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

 

[Nexus99]

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

So cos instead of sin?

 

[haruspex]

So cos instead of sin?

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.

 

[Nexus99]

Can the

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?

 

[etotheipi]

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 $a'_{2,x'} = a_x \cos{\beta}$

 

[Nexus99]

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]

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

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.