Calculate the magnitude of driving force on the car

[chwala]

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...

 

[Doc Al]

How many forces act on the car?

 

[Doc Al]

When you apply Newton's 2nd law, take all the forces into account.

 

[Lnewqban]

 

[chwala]

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.

 

[MatinSAR]

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 $mg$.

 

[chwala]

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 ...

 

[MatinSAR]

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.

In your solution $F_0=mgsin \theta$.

 

[chwala]

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

 

[MatinSAR]

aaaargh i think i've seen it.

$\cos 72^0 = \dfrac{mg}{8500}$ therefore $mg = 1767.25$ N

No. Rethink.

 

[MatinSAR]

mg has 2 components. One perpendicular to the surface and one parallel to the surface.
$mgcos 72$ is one of the components.

 

[chwala]

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

 

[MatinSAR]

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:

If there is any problem I'll be happy to help

 

[MatinSAR]

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 $R$.

 

[chwala]

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...

 

[chwala]

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!