# Law of Motion (forces)

The air exerts a forward force of 10 N on the propeller of a 0.20-kg model airplane. if the plane accelerates forward at 2.0 m/s², what is the magnitude of the resistive force exerted by the air on the airplane?

A boat moves through the water with two forces acting on it. ne is a 2000 N forward push by the motor, and the other is an 1800-N resistive force due to the water. a) what is the acceleration of the 1000 kg boat? b) if it starts frpù rest, how far will it move in 10 s? c) what will its velocity be at the end of this time?

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Hootenanny
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Please use the template provided and provide some details of how you have attempted the problems. Since you have enough knowledge to know that these are Laws of Motion problems involving forces, I'm sure you could at least hazard a guess as to how you should approach the problems.

I was looking at the other threads! and yes they use a template which somehow does not appear on my screen when I post a thread....
I'm assuming for the first question you have to add up the Y forces and the x forces since when we draw the diagram there is one vector facing upward, and the second vector is moving 90 degrees east!
In doing so, you find the resultant force, but here they're asking for the resistive force! Plus, we're given velocity of the object and not the magnitudes of its force vectors! I'm still confused!

Hootenanny
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I was looking at the other threads! and yes they use a template which somehow does not appear on my screen when I post a thread....
I'm guessing you have the nexus skin, the template isn't implemented in that skin yet (and I dont' think it will be :grumpy: )
I'm assuming for the first question you have to add up the Y forces and the x forces since when we draw the diagram there is one vector facing upward, and the second vector is moving 90 degrees east!
In doing so, you find the resultant force, but here they're asking for the resistive force! Plus, we're given velocity of the object and not the magnitudes of its force vectors! I'm still confused!
Okay, I think that your analysing the problem wrong. In the question we should assume the plane is in level flight, i.e. there is no net force in the vertical direction. Now, the propellers exert a force forwards and the resistive force acts backwards (against the direction of motion). Note also that you are given an acceleration not a velocity.

would the equation F=ma be of some help in this situation?
where m= 0.20 kg and a 2.0m/s²....
hence, the weight or the gravitational force is 0.4 N ....do i now try to find the resultant force? i'm assuming when you use the parallegram method to add the vectors, the resultant force (the resistive force) will be in the opposite direction of the forward force?????!

Hootenanny
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would the equation F=ma be of some help in this situation?
Good, with one small modification; $\sum \vec{F}=m\vec{a}$. I.e. the net force
where m= 0.20 kg and a 2.0m/s²....
Again, good.
hence, the weight or the gravitational force is 0.4 N ....do i now try to find the resultant force? i'm assuming when you use the parallegram method to add the vectors, the resultant force (the resistive force) will be in the opposite direction of the forward force?????!
Almost. I will refer you to my previous post;
me said:
In the question we should assume the plane is in level flight, i.e. there is no net force in the vertical direction. Now, the propellers exert a force forwards and the resistive force acts backwards (against the direction of motion).
This means we can ignore any forces which act in the vertical plane (i.e. gravity and lift). All we need to consider is forces acting in the horizontal plane, i.e. thrust from the propeller (forward) and the resistive force from the air (backwards). Do you follow?

Yes i do, but my only concern is that the backward and forward forces are along the same line....how can you find the net force? and what does 0.4 N represent if you say there is no net force in the vertical direction.
Thank you, i appreciate the help.

i know that the forward and backward forces equal to zero according to Newton's law....
Right?

Hootenanny
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Yes i do, but my only concern is that the backward and forward forces are along the same line....how can you find the net force?
Okay, me and you are stood on opposite sides of a large box which rests on the floor. Now, we both push on the box in opposite directions (say I push to the left and you push to the right). I push with a force of 10N, you push with a force of 20N. What is the resultant/net force?
and what does 0.4 N represent if you say there is no net force in the vertical direction.
The 0.4N is the weight of the aircraft acting downwards. However, since the plane is in level flight (the question does not state this but should do :grumpy: ) this weight is perfectly balanced by the lift (upwards) created by the wings; so there is no net force in the vertical plane.

Hootenanny
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i know that the forward and backward forces equal to zero according to Newton's law....
Right?
Why do you say that?

No it's not zero since acceleration is not zero...in response to your question, i subtract my force by your's!
10-resistive force= net force ? we have two unknowns??

Hootenanny
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No it's not zero since acceleration is not zero...in response to your question, i subtract my force by your's!
Correct!
10-resistive force= net force ? we have two unknowns??
Yes, but what is the net force equal to?

is it 0.4?
resistive force= -(-10+0.4)
9.6 N?

Hootenanny
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is it 0.4N?
Correct!

And just to correct one of my previous posts; the 0.4N doesn't represent the weight of the aircraft, my apologies. The weight of the aircraft is 0.2x9.81=1.962. A careless error on my part (although in my defence is 01:20 AM in the UK :zzz: )

Last edited:
thank you! whao! it took me over five posts to realize it was all about subtracting the vectors! silly me!

Hootenanny
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thank you! whao! it took me over five posts to realize it was all about subtracting the vectors! silly me!
My pleasure Now, for your second question part (a) is very similar to the question you just answered (but the opposite way round). Parts (b) and (c) can be answered using kinematic equations.

As for me, I'm off to bed :zzz: but I'm sure plenty of others will give you a hand around here, if you need it.

hahaha, it's the same time in morocco!

all right, good night, thank you again!

Hootenanny
Staff Emeritus