Forces and Acceleration of a Spacecraft

[olivia ivaldi]

Homework Statement

A spacecraft has two adjustable engines that exert the same force.

When the engines push in the same direction and in the same sense, the spacecraft takes 28 seconds to travel a certain distance, starting from rest.

Then, the engines are positioned so that the two forces are perpendicular to each other.

Relevant Equations

Calculate how long the spacecraft takes to travel the same distance as before.

My attempt and thoughts
I know that the two engines exert the same force and that the spacecraft starts from rest.

First, I need to consider the resultant force when the two engines push in the same direction. Then, I need to calculate the resultant force when they are perpendicular.

I think I should use Newton’s second law ((F=ma)) to find the acceleration in both situations.

Since the distance is the same in both cases and the initial velocity is zero, I can use the equations of uniformly accelerated motion to compare the two times.
My question

How can I compare the two accelerations to calculate the new time needed to cover the same distance?

 

[PeroK]

Force and acceleration are vectors. Technically, $a = |\vec a|$ is the magnitude of the acceleration.

Does that help?

 

[olivia ivaldi]

Force and acceleration are vectors. Technically, $a = |\vec a|$ is the magnitude of the acceleration.

Does that help?

I understand that force and acceleration are vectors, but I still don’t understand how to relate the two accelerations to find the final time.

 

[PeroK]

I understand that force and acceleration are vectors, but I still don’t understand how to relate the two accelerations to find the final time.

Do you understand the concept of vector addition?

 

[olivia ivaldi]

Do you understand the concept of vector addition?

Yes

 

[PeroK]

Yes

What does it tell you in this case?

 

[olivia ivaldi]

What does it tell you in this case?

In this case, when the two forces act in the same direction, the resultant force is 2F. When they are perpendicular, the resultant force is √2 F, using the Pythagorean theorem.

 

[PeroK]

In this case, when the two forces act in the same direction, the resultant force is 2F. When they are perpendicular, the resultant force is √2 F, using the Pythagorean theorem.

What does that tell you about the magnitude of the acceleration in the two cases?

 

[PeroK]

Perhaps call them $a_1$ and $a_2$.

 

[berkeman]

Not to overly-complicate the problem, but does it make a difference that the engine forces are not applied at the Center of Gravity (CoG) of the spaceship?

 

[olivia ivaldi]

Perhaps call them $a_1$ and $a_2$.

Oh, I understand now! Since the mass is the same in both cases, the acceleration is proportional to the resultant force. Therefore, a1 = √2 · a2. Thank you for your help!

 

[olivia ivaldi]

Not to overly-complicate the problem, but does it make a difference that the engine forces are not applied at the Center of Gravity (CoG) of the spaceship?

I hadn’t thought about that. I just followed the information given in the exercise and calculated the resultant forces. Thank you for pointing that out!

 

[PeroK]

Not to overly-complicate the problem, but does it make a difference that the engine forces are not applied at the Center of Gravity (CoG) of the spaceship?

I've noticed that in the past few years, more and more questions are written by people who appear to have no grasp of the physics. This is a typical example, where they are oblivious to the additional torque.

 

[olivia ivaldi]

I've noticed that in the past few years, more and more questions are written by people who appear to have no grasp of the physics. This is a typical example, where they are oblivious to the additional torque.

I appreciate your feedback. I’m a 16-year-old student in my third year of high school in Italy, so I’m still learning. I’m studying physics in Italian, and I’m using this forum, which was recommended by my teacher, specifically to ask questions like this one. Thank you for your patience and understanding!

 

[PeroK]

I appreciate your feedback. I’m a 16-year-old student in my third year of high school in Italy, so I’m still learning. I’m studying physics in Italian, and I’m using this forum, which was recommended by my teacher, specifically to ask questions like this one. Thank you for your patience and understanding!

I'm taking about the people who write the textbooks! I wonder who reviews these questions before publication?

 

[olivia ivaldi]

I'm taking about the people who write the textbooks! I wonder who reviews these questions before publication?

You and me both. thank you for the clarification.

 

[erobz]

Not to overly-complicate the problem, ...

analytically intractable!

 

[A.T.]

I've noticed that in the past few years, more and more questions are written by people who appear to have no grasp of the physics. This is a typical example, where they are oblivious to the additional torque.

The question text by itself is one thing, but it's the illustration (usually made by an artist) that strongly suggests a non-zero net torque.

I hadn’t thought about that. I just followed the information given in the exercise and calculated the resultant forces.

With no further information information provided, that's probably what you were expected to do.

But as you can maybe imagine, the 2nd rocket would start spinning and would follow a spiraling path, and in general it would not be aligned with the velocity vector. In such a situation, part of the acceleration (component perpendicular to the current velocity vector) acts just to change the velocity direction. So you cannot use the simple formula for distance under constant acceleration along a straight path anymore, because the tangential component of acceleration (parallel to velocity) is varying.

 

[DaveC426913]

With no further information information provided, that's probably what you were expected to do.

And this might lead to bonus points with your teacher, to-wit:

"If the engines are not arranged around the CoM, they will generate a torque, rendering the problem intractable as-stated. If they are arranged so they do not generate a torque, the answer is x."

Many such puzzles can be answered (qualitatively, if not quantitatively) as along as you state your assumptions.

A good teacher should reward you for mentioning such assumptions, provided you also give your best answer to the question. (But don't over-use this, you don't want to come across as a smart-ass).

 

[FranzDiCoccio]

The question text by itself is one thing, but it's the illustration (usually made by an artist) that strongly suggests a non-zero net torque.

This problem is actually Problem 17 in Chapter 4 of Physics, 9th edition, by Cutnell et al., from which the Italian textbook draws much of its material.

In this problem (and throughout that chapter of the Italian textbook), all objects are treated as point masses, so torque is not an issue. Equivalently, as has already been suggested, one must assume that all forces act through the body's center of mass.

The same applies to the exercises in Chapter 4 of the original textbook. In fact, the original wording of Problem 17 does not even bother to restate this assumption.

I do agree that the added illustration (which is not present in the original problem) obscures the problem rather than clarifying it, and that it definitely warrants an erratum. I've often found that errors in the Italian textbook arise from additions to the original material.

Students at this stage of their education don't have the theoretical background to deal with torque yet. This topic will be formally introduced in a few months. However, a perceptive student might well notice that something is wrong with the illustration, without having the tools to explain exactly what. I agree that bringing up such questions in class could lead to a valuable discussion.

 

[berkeman]

In this problem (and throughout that chapter of the Italian textbook), all objects are treated as point masses, so torque is not an issue.

Thanks very much for the clarification, Franz!