# Problem with torque, angular momentum and forces

• Granger
In summary, the problem is to determine the maximum allowable acceleration of a truck, with a 1.8m board placed inside, in order for the board to remain in a certain position. The forces acting on the board are its weight and the normal reactions from the vertical partition and the block secured to the floor. Additional equations are needed to solve for the unknowns (N, a, and m), using the reference point of the bottom block for angular momentum and torques. However, when attempting to solve the system of equations, the normal reaction was found to be zero, leading to an impossible solution. Further clarification is needed on the angle and the forces acting on the board.
Granger
Homework Help Template added by Mentor

## 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!

Last edited by a moderator:
Anybody?

Granger said:

## 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.

Yes you're right, here it is:

Granger said:
the normal reactions that the vertical partition and the block exert on the body (which are equal).
Why would they be equal?

Granger said:

## 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) ?

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.

haruspex said:
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).

scottdave said:
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?

Granger said:
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.

haruspex said:
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.

Granger said:
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!

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!

## 1. What is torque and how is it related to angular momentum?

Torque is the measure of the force that causes an object to rotate around an axis. It is directly related to angular momentum, which is the measure of an object's rotational motion. Torque is the cause of changes in angular momentum, and the greater the torque applied, the greater the change in angular momentum.

## 2. How can I calculate torque?

Torque can be calculated by multiplying the force applied to an object by the distance between the point of application of the force and the axis of rotation. This can be represented by the equation τ = F x r, where τ is torque, F is force, and r is distance.

## 3. What is the difference between angular and linear momentum?

Angular momentum is the measure of an object's rotational motion, while linear momentum is the measure of an object's linear motion. Angular momentum is affected by the object's mass, speed, and distance from the axis of rotation, while linear momentum is affected by an object's mass and velocity.

## 4. How do forces affect torque and angular momentum?

Forces can cause a change in an object's torque and angular momentum. If a force is applied at a distance from the axis of rotation, it will create torque and cause the object to rotate. The direction of the force also determines the direction of the rotation and the change in angular momentum.

## 5. What are some real-life applications of torque and angular momentum?

Torque and angular momentum have many applications in everyday life, such as in the operation of machines and vehicles, like cars and bicycles. They are also important in sports, such as in the spin of a ball in baseball or the rotation of a figure skater. In addition, understanding torque and angular momentum is crucial in the design and functioning of tools and equipment, such as wrenches and gears.

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