Calculating Acceleration and Velocity of a Boat in Motion

I mean, if I asked you to sum all the forces acting on the box in the horizontal plane (so you are only interested in forces to the left or right) what would that sum be equal to?
  • #1
L²Cc
149
0
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?

Please help me solve these questions; I do not know what method to use. SOS!
 
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  • #2
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.
 
  • #3
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!
 
  • #4
L²Cc said:
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: )
L²Cc said:
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.
 
  • #5
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?!
 
  • #6
L²Cc said:
would the equation F=ma be of some help in this situation?
Good, with one small modification; [itex]\sum \vec{F}=m\vec{a}[/itex]. I.e. the net force
L²Cc said:
where m= 0.20 kg and a 2.0m/s²...
Again, good.
L²Cc said:
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?
 
  • #7
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.
 
  • #8
i know that the forward and backward forces equal to zero according to Newton's law...
Right?
 
  • #9
L²Cc said:
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?
L²Cc said:
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.
 
  • #10
L²Cc said:
i know that the forward and backward forces equal to zero according to Newton's law...
Right?
Why do you say that?
 
  • #11
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??
 
  • #12
L²Cc said:
No it's not zero since acceleration is not zero...in response to your question, i subtract my force by your's!
Correct!
L²Cc said:
10-resistive force= net force ? we have two unknowns??
Yes, but what is the net force equal to?
 
  • #13
is it 0.4?
resistive force= -(-10+0.4)
9.6 N?
 
  • #14
L²Cc said:
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:
  • #15
thank you! whao! it took me over five posts to realize it was all about subtracting the vectors! silly me!
 
  • #16
L²Cc said:
thank you! whao! it took me over five posts to realize it was all about subtracting the vectors! silly me!
My pleasure :smile: 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.
 
  • #17
hahaha, it's the same time in morocco!
 
  • #18
all right, good night, thank you again!
 
  • #19
L²Cc said:
hahaha, it's the same time in morocco!
عليكم ان عمل جنوني في هذه اللحظة. غودنيت

Vous devez être fonctionnement fou à cette heure. Bonne nuit
 
  • #20
Je vais bientot dormir ! moi aussi je suis fatiqué. Je vous remercie pour votre aide!
 

1. What is Newton's First Law of Motion?

The first law of motion, also known as the Law of Inertia, states that an object at rest will stay at rest and an object in motion will stay in motion at a constant velocity unless acted upon by an external force.

2. What does Newton's Second Law of Motion state?

Newtons's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. This can be represented by the equation F=ma, where F is force, m is mass, and a is acceleration.

3. How does Newton's Third Law of Motion apply to everyday life?

Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This can be observed in everyday life, such as when pushing a shopping cart, the force applied to the cart causes the cart to move forward while also exerting an equal and opposite force on the person pushing it.

4. Can the Law of Motion be applied to objects in space?

Yes, the Law of Motion applies to objects in space. In fact, it was Newton's laws of motion that helped explain the motion of planets and other celestial bodies in our solar system.

5. What are some real-life examples of the Law of Motion?

Some real-life examples of the Law of Motion include a ball rolling down a hill, a car accelerating when the gas pedal is pressed, a person jumping off a diving board, and a rocket launching into space.

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