Problem with torque, angular momentum and forces

[Granger]

Homework Statement

I have the following problem to solve:

A 1.8m board is placed in a truck with one end resting against a block secured to the floor and the other one leaning against a vertical partition. The angle the Determine the maximum allowable acceleration of die truck if the board is to remain in the position shown.

If you put this problem on google you can find an image (if it helps). The truck moves from left to right.

The Attempt at a Solution

So I first began to thought that both velocity and acceleration of the board are directed to the right.
The forces acting on the body are its weight, and the normal reactions that the vertical partition and the block exert on the body (which are equal).

Then putting this on equations:

x direction: $$ ma_x=N\cos\theta-N$$
and y direction: $$0=N\sin\theta -mg$$

We have 2 equations and 3 unknowns (N, a and m).
We need a 3rd equation which is

$\frac{d}{dt}L_{system}=\sum(\tau_{net})$
(these are supposed to be vectors)

And so if we choose the bottom block as reference point to gives us angular momentum and torques we have (and this is the equation I'm not sure about)

$$ -m\frac{l}{2}a\sin \theta= lN\sin(105) - \frac{l}{2}mg\sin(165)$$

(the plus and minus sign appear because of the direction of torque and the direction of angular momentum are given by the right hand rule for cross products).

This leading me to a system of 3 equations.
However if I try to solve this system (for example, isolate mass in eq(1) and substitute in eq (2) I end up with N=0 and therefore m=0 which is absurd). Can someone help me figuring out this problem?

Thanks!

 

[Granger]

Anybody?

 

[haruspex]

Homework Statement

I have the following problem to solve:

A 1.8m board is placed in a truck with one end resting against a block secured to the floor and the other one leaning against a vertical partition. The angle the Determine the maximum allowable acceleration of die truck if the board is to remain in the position shown.

If you put this problem on google you can find an image (if it helps). The truck moves from left to right.

The Attempt at a Solution

So I first began to thought that both velocity and acceleration of the board are directed to the right.
The forces acting on the body are its weight, and the normal reactions that the vertical partition and the block exert on the body (which are equal).

Then putting this on equations:

x direction: $$ ma_x=N\cos\theta-N$$
and y direction: $$0=N\sin\theta -mg$$

We have 2 equations and 3 unknowns (N, a and m).
We need a 3rd equation which is

$\frac{d}{dt}L_{system}=\sum(\tau_{net})$
(these are supposed to be vectors)

And so if we choose the bottom block as reference point to gives us angular momentum and torques we have (and this is the equation I'm not sure about)

$$ -m\frac{l}{2}a\sin \theta= lN\sin(105) - \frac{l}{2}mg\sin(165)$$

(the plus and minus sign appear because of the direction of torque and the direction of angular momentum are given by the right hand rule for cross products).

This leading me to a system of 3 equations.
However if I try to solve this system (for example, isolate mass in eq(1) and substitute in eq (2) I end up with N=0 and therefore m=0 which is absurd). Can someone help me figuring out this problem?

Thanks!

Your description of the set up is garbled and rather unclear. If you found an image on the net, please post the link.

 

[Granger]

Yes you're right, here it is:

 

[haruspex]

the normal reactions that the vertical partition and the block exert on the body (which are equal).

Why would they be equal?

 

[scottdave]

Homework Statement

... against a vertical partition. The angle the Determine the maximum allowable acceleration...
...we choose the bottom block as reference point to gives us angular momentum and torques we have (and this is the equation I'm not sure about)

$$ -m\frac{l}{2}a\sin \theta= lN\sin(105) - \frac{l}{2}mg\sin(165)$$

(the plus and minus sign appear because of the direction of torque and the direction of angular momentum are given by the right hand rule for cross products).

This leading me to a system of 3 equations.
However if I try to solve this system (for example, isolate mass in eq(1) and substitute in eq (2) I end up with N=0 and therefore m=0 which is absurd). Can someone help me figuring out this problem?

In your initial problem statement, some info got left out " The angle the Determine the maximum " What is the angle? Where did you get the 105 and 165 degrees in your sin(105) and sin(165) ?

 

[scottdave]

What if you approached it like this: Pretend the forward partition is not, there and find the necessary acceleration in order to keep the board at the desired angle.
If the board is hovering in relation to the truck, then the center of mass of the board is accelerating at the same rate as the truck.
This essentially the condition at just slightly faster acceleration, when the board just starts to lift away from the partition.

 

[Granger]

Why would they be equal?

I thought they were equal because the table was fixed (so they equilibrate each other). Thinking more about it this is probably wrong because all the 3 forces (weight, normal reaction and force of the truck).

In your initial problem statement, some info got left out " The angle the Determine the maximum " What is the angle? Where did you get the 105 and 165 degrees in your sin(105) and sin(165) ?

You're right I forgot to write the angle. It should say: The angle the board makes with the base of the truck. Determine the maximum allowable acceleration of die truck if the board is to remain in the position shown.

I didn't quite understand your approach, sorry. Can you elaborate just a bit more, please?

 

[haruspex]

this is probably wrong because all the 3 forces (weight, normal reaction and force of the truck).

Not sure what you mean by "force of the truck".
The weight and the normal reaction from the vertical surface are two forces, yes.
I interpret the question as saying the bottom of the board rests on the floor of the truck (a vertical normal force) and against a block that is also on the floor (providing a second horizontal normal force). You can combine hose as a single force if you prefer, but its line of action need not be at the same angle as the board. Better to treat them as two separate forces.

And these forces will not be in balance - there is an acceleration.

 

[Granger]

Not sure what you mean by "force of the truck".
The weight and the normal reaction from the vertical surface are two forces, yes.
I interpret the question as saying the bottom of the board rests on the floor of the truck (a vertical normal force) and against a block that is also on the floor (providing a second horizontal normal force). You can combine hose as a single force if you prefer, but its line of action need not be at the same angle as the board. Better to treat them as two separate forces.

And these forces will not be in balance - there is an acceleration.

Oh ok! I thought it was just a single normal reaction.
I think I'm having trouble relating all the forces. I'm simply not understanding the boundary condition and how to relate the torques.

 

[haruspex]

Oh ok! I thought it was just a single normal reaction.
I think I'm having trouble relating all the forces. I'm simply not understanding the boundary condition and how to relate the torques.

Assign a unique label to each force.
Write the ΣF=ma equation for the horizontal direction. (The vertical direction is trivial.)
Take moments about the board's mass centre and write the Στ=Iα equation (sum of torques = moment of inertia x angular acceleration).
Note that because there is a linear acceleration you must use the mass centre as the axis for torque balance!

 

[Granger]

Hi! I was finally able the solve the problem.
The equation that relates the sum of torques and the derivative of the angular momentum is enough to calculate the acceleration.
What I was doing wrong was considering the torque of the normal reaction in B which is zero in the boundary condition of the board falling (loses contact in B).

Thank you all for your contributions to help me understand the problem!