Calculate the magnitude of driving force on the car

  • #1

chwala

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Homework Statement
A car of mass ##850## kg is travelling with acceleration ##0.3 m/s^2##, up a straight road inclined at ##12^0## to the horizontal. There is a force resisting motion of ##250## N. Calculate the magnitude of the driving force.
Relevant Equations
Resolving forces- Mechanics
My approach is as follows, let ##D## be driving force and ##F## the force resisting motion, then

##D-F = 850 × 0.3##

##D = 250 + 255##

##D = 505##

Also, Force parallel to the road is given by, ##F_1 = 8500 \cos 78^0 =1767.24 N##

Therefore, the magnitude of Driving force is given by, ##F_1 + D = 2272.25 N##

your insight is welcome...not sure on this...
 
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  • #2
How many forces act on the car?
 
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  • #3
When you apply Newton's 2nd law, take all the forces into account.
 
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  • #4
Car on hill.jpg
 
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  • #5
Since we are interested in the driving force then, before motion ( that is at equilibrium ) ##a=0##

##F_0 - 8500 \cos 78^0 = 0##

##F_0=1767.25## N


we know that the Force resisting motion ( car now in motion)= 250 N. Therefore,


##F_1 - 250= ma ##

##F_1= (850 × 0.3 )+250##

##F_1= 505## N

Therefore,

Driving force ##D=F_0 +F_1 = 1767.25+ 505= 2272.25## N

Of course i could be wrong...it is a new area for me ....cheers.
 
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  • #6
chwala said:
Since we are interested in the driving force then, before motion ( that is at equilibrium ) ##a=0##

##F_0 - 8500 \cos 78^0 = 0##

##F_0=1767.25## N


we know that the Force resisting motion ( car now in motion)= 250 N. Therefore,


##F_1 - 250= ma ##

##F_1= (850 × 0.3 )+250##

##F_1= 505## N

Therefore,

Driving force ##D=F_0 +F_1 = 1767.25+ 505= 2272.25## N

Of course i could be wrong...it is a new area for me ....cheers.
I agree with final answer but do you know what ##F_0## is?
If you ask me It is better to use post #4 to identify net force acting on car in terms of ##D## and ##F## and ##mg##.
 
  • #7
MatinSAR said:
I agree with final answer but do you know what ##F_0## is?
That should be the vertical force i.e acting perpendiculor to the road. ...let me check on this ..what i have on my textbook notes is a bit confusing...see the attachment ...
 

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  • #8
chwala said:
That should be the vertical force i.e acting perpendiculor to the road. ...let me check on this ..what i have on my textbook notes is a bit confusing...see the attachment ...
But what causes this force? It's interesting that your book doesn't mention it! Look at digram in post #4. It's the gravitational force.
2024_02_12 1_23 PM Office Lens.jpg


In your solution ##F_0=mgsin \theta ##.
 
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  • #9
MatinSAR said:
But what causes this force? It's interesting that your book doesn't mention it! Look at digram in post #4. It's the gravitational force.
View attachment 340203
aaaargh i think i've seen it.

##\cos 72^0 = \dfrac{mg}{8500}## therefore ##mg = 1767.25## N
 
  • #10
chwala said:
aaaargh i think i've seen it.

##\cos 72^0 = \dfrac{mg}{8500}## therefore ##mg = 1767.25## N
No. Rethink.
 
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  • #11
mg has 2 components. One perpendicular to the surface and one parallel to the surface.
##mgcos 72 ## is one of the components.
 
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  • #12
I can have my equation as,

##D= R+F+ma##

and noting that

##[ \cos 78^0 = \sin 12^0]##

then,

##D= mg \cos 78^0 + 250+ 255 = 2272.25##N
 
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  • #13
chwala said:
I think in my book ##mg## is what they are calling ##N## and ##R-N=0## and ##N## using the right angle triangle is given by ##\cos 72^0 = \dfrac{N}{8500}##.

Do confirm again on this.
Your book doesn't use ##N##. It used ##R## to show the normal force.
Let's solve the problems one by one. For question in post #7:
2024_02_12 1_57 PM Office Lens.jpg

If there is any problem I'll be happy to help
 
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  • #14
chwala said:
I can have my equation as,

##D= R+F+ma##

and noting that

##[ \cos 78^0 = \sin 12^0]##

then,

##D= mg \cos 78^0 + 250+ 255 = 2272.25##N
I guess you're answering question in post #1 now. Is everything clear with question in post #7?
What is ##R## in your solution? If it is the normal force then you are wrong. See post #8 and note that I've showed normal force using ## N## instead of ##R##.
 
  • #15
MatinSAR said:
I guess you're answering question in post #1 now. Is everything clear with question in post #7?
What is ##R## in your solution? If it is the normal force then you are wrong. See post #8 and note that I've showed normal force using ## N## instead of ##R##.
... Let me check and go through the literature... will get back in a few hours...
 
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  • #16
MatinSAR said:
Your book doesn't use ##N##. It used ##R## to show the normal force.
Let's solve the problems one by one. For question in post #7:
View attachment 340206
If there is any problem I'll be happy to help
This part is now clear; i was mixing up the two forces...now using my attachment in post ##7## as reference, it is clear that the force acting parallel to the ramp is given by:
Firstly, ##F=ma## since the crate is at rest then ##a=0##.
now
There are two opposing forces that will enable equilibrium, let me have them as,
##F_1 - F_0 =0## where ##F_1## is the frictional force. ##F_1## makes an angle ##θ## with the parallel direction ##F_0## where ##F_0## using pythagoras theorem is given by ##F_0=300 \cos 72^0## or ##F_0 = 300\sin 18^0##.
Similarly, the force acting perpendicularly to the ramp is given by:
##R -N=0## where ##R## is the Normal contact force and ##N## is the force acting perpendicularly to ramp,
##N=300 \cos 18^0## using the pythagoras theorem. From here the steps to solution are straightforward.

Now to my problem, using Newton's second law of motion, i shall have,

##D-F_1 -F_0 =ma##

##D-250-8500 \cos 78^0 = 850×0.3## or

##D-250-8500 \sin 12^0 = 850×0.3##

##D=250+1767.25+255##

##D=2272.25##N.

Cheers man!
 
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