B Fundamental Problems With Exam Question: Moving Truck & Acceleration

[neilparker62]

TL;DR Summary

Badly worded (at least) exam question

Is it just me or are there some fundamental problems with this exam question ? What is driving me bananas is you have a moving truck (constant velocity). Then the truck "accelerates" but the block in the back of the truck "stays in the same place". Does this mean relative to the ground, relative to the "constant velocity" truck frame or relative to the now accelerating truck frame ? How do you draw a free body diagram for an object in an accelerating frame ?

 

[DrClaude]

Then the truck "accelerates" but the block in the back of the truck "stays in the same place". Does this mean relative to the ground, relative to the "constant velocity" truck frame or relative to the now accelerating truck frame ?

I read it as relative to the truck itself.

 

[PeroK]

Does this mean relative to the ground, relative to the "constant velocity" truck frame or relative to the now accelerating truck frame ?

It must mean relative to the truck. You can infer this from the following parts of the question

How do you draw a free body diagram for an object in an accelerating frame ?

A frame is essentially a system of coordinates. An object is not in any frame; an object is in all frames. In this case, the crate is accelerating with the truck and a free-body diagram shows the forces that cause that acceleration.

 

[neilparker62]

Ok - many thanks all. "Just me" indeed I guess. But what about 4.2.5. Acceleration relative to what ?

 

[anorlunda]

Ok - many thanks all. "Just me" indeed I guess. But what about 4.2.5. Acceleration relative to what ?

relative to the truck.

 

[jbriggs444]

relative to the truck.

I disagree with this assessment. It must be relative to the ground because the acceleration of the truck in 4.2.5 is unspecified and could be arbitrarily large. As a result, the acceleration of the crate relative to the truck will be impossible to determine. But the acceleration of the crate relative to the ground is easy to calculate.

Edit: I do see a minor concern with the wording of 4.2.2

We are asked to graphically depict the direction of motion of the truck but not the direction of its acceleration. Technically the two need not be aligned. The questioner presumably assumes the forward acceleration of a forward-moving truck along a straight highway.

 

[PeroK]

relative to the truck.

I agree with @jbriggs444. I'd assume we are talking about (real) acceleration relative to the inertial ground frame.

 

[neilparker62]

I disagree with this assessment. It must be relative to the ground because the acceleration of the truck in 4.2.5 is unspecified and could be arbitrarily large. As a result, the acceleration of the crate relative to the truck will be impossible to determine. But the acceleration of the crate relative to the ground is easy to calculate.

I am borrowing from this "cheat sheet / video".

A maximum possible acceleration for the truck can be determined. Then I guess the difference between that and the crate's acceleration relative to ground would give the (backward) acceleration) of the crate relative to the truck in the event of the crate slipping. Assuming a more or less instant 'transition' from static to kinetic friction.

 

[PeroK]

I am borrowing from this "cheat sheet / video".

View attachment 316086

A maximum possible acceleration for the truck can be determined. Then I guess the difference between that and the crate's acceleration relative to ground would give the (backward) acceleration) of the crate relative to the truck in the event of the crate slipping. Assuming a more or less instant 'transition' from static to kinetic friction.

Yes, in general the maximum acceleration due to static fraction in a case like this is $\mu_sg$. If the acceleration of the truck exceeds this, then the crate slips and the acceleration of the crate reduces to $\mu_kg$.