1. ## Mechanics help!

1) A particle of mass 9kg is sliding down a smooth inclined plane with an acceleration of 4.9m per second per second. Find the angle of inclination of the plane.

2) A parcel of mass 60kg is released from rest on a smooth plane inclined at arcsin3/5 to the horizontal. Find the velocity of the parcel when it has travelled 5m down the plane.

3) A horizontal force of 2N is just sufficient to prevent a block of mass 1kg from sliding down a rough plane inclined at arcsin 7/25 to the horizontal. Find the coefficient of friction between the block and the plane and the acceleration with which the block will move when the force is removed.

4) A stone slides in a straight line across a frozen pond. Given that the initial speed of the stone is 5m per second and that it slides 20m before coming to rest, calculate the coefficient of friction between the stone and the surface of the frozen pond.

Any help would be much appreciated. Thanks in advance.

2. Originally Posted by Confuzzled?
1) A particle of mass 9kg is sliding down a smooth inclined plane with an acceleration of 4.9m per second per second. Find the angle of inclination of the plane.
I am not a big mechanic/physics guy so I might be totally wrong but I have a reasonable guess for this one. The fact that is is 9kg is irrelevent. What is relevent that the acceleration on an inclined plance is 4.9m/s2. While regular acceleration due to gravity is 9.8m/s2. Thus, it is reasonable to say it is half the angle of free fall. Since free fall happens at 90 degrees this one happens at 45 degrees.

3. Originally Posted by Confuzzled?
1) A particle of mass 9kg is sliding down a smooth inclined plane with an acceleration of 4.9m per second per second. Find the angle of inclination of the plane.
As the plane is smooth the only forces acting on the particle are the weight
and reaction force between the plane and particle. We resolve these into
components down the plane and normal to the plane.

Now the reaction force has no component down the plane, while the weight
has a component $mg \sin (\theta)$ down the plane, so down the plane we have:

$
ma=mg \sin(\theta)
$

where $a$ is the acceleration down the plane. So

$
4.9=9.8 \sin(\theta)
$
,

or:

$
\sin(\theta)=1/2
$
.

But by definition $\theta$ is in the first quadrant so:

$
\theta=\arcsin(1/2)=30^{\circ}$
, or $\pi/6$ radians

(the component of the gravitational force normal to the plane cancels
out the normal reaction and so neither of these need be considered further
in this problem)

RonL

4. Originally Posted by Confuzzled?
2) A parcel of mass 60kg is released from rest on a smooth plane inclined at arcsin3/5 to the horizontal. Find the velocity of the parcel when it has travelled 5m down the plane.
From the solution to question 1), we see that the accelaration down the
plane is:

$
a=g \sin(\theta)
$
,

so in this case $\sin(\theta)=3/5$ so:

$
a=9.8 \times (3/5)=5.88\ \mbox{m/s}
$

RonL

5. Originally Posted by Confuzzled?
3) A horizontal force of 2N is just sufficient to prevent a block of mass 1kg from sliding down a rough plane inclined at arcsin 7/25 to the horizontal. Find the coefficient of friction between the block and the plane and the acceleration with which the block will move when the force is removed.
If I have done this right:

As before we resolve the gravitational force on the particle into components
down the plane and normal to the plane.

The component down the plane is $mg \sin(\theta)$ and that
normal to the plane is $-mg \cos(\theta)$

We also have a horizontal force on the particle which we also resolve into
components down the plane and normal to the plane.

The component down the plane is $-2\cos(\theta)$ (the negative sign here
is due to this force acting in the horizontal direction pointing up the plane)
and normal to the plane is $-2 \sin(\theta)$

The remaining forces on the partice are the normal reaction which acts
normaly to the plane and just balance the other forces normal to the plane,
so: $N=mg \cos(\theta)+2 \sin(\theta)$, and the frictional
force which is just balancing the forces down the plane:

$F=\mu R=mg \sin(\theta)-2\cos(\theta)$

So:

$
\mu =\frac{mg \sin(\theta)-2\cos(\theta)}{mg \cos(\theta)+2 \sin(\theta)}\approx0.0827
$

Some of the relevant forces are shown in the attachment.

RonL

6. Originally Posted by Confuzzled?
4) A stone slides in a straight line across a frozen pond. Given that the initial speed of the stone is 5m per second and that it slides 20m before coming to rest, calculate the coefficient of friction between the stone and the surface of the frozen pond.
The only effective force acting on the stone in the horizontal plane is friction:

$
F=-\mu N
$

where $\mu$ is the coefficient of friction and $N$
is the normal contact force, which in this case is $mg$.

Now:

$
\dot{v}=F/m=-\mu g
$

Integrating this gives:

$
v=- \mu g t + v_0=-\mu g t +5
$

and integrating again:

$
s=-\mu g t^2/2 + 5t +s_0=-\mu g t^2/2 + 5t
$

if we take $s_0=0$.

Now when the stone comes to rest $v=0$, this occurs when:

$
t=\frac{5}{\mu g}
$

and at this time $s=20$, so we have:

$
20=-\mu g t^2/2 + 5t =-\frac{25}{2 \mu g}+\frac{25}{\mu g}=\frac{25}{2 \mu g}
$

which gives:

$
\mu=\frac{25}{40 g}\approx 0.064
$

RonL

7. thank you so much for your help but question 2 asks for velocity.

8. Originally Posted by Confuzzled?
thank you so much for your help but question 2 asks for velocity.
Integrate the acceleration once it gives you the velocity (initial velocity is
zero), integrate again to give distance moved (again initial distance moved is zero).

Put distance equal to 5 and solve for t, then substitute this t back into the
expression for velocity.

RonL

9. Hi:

One approach is to equate kinetic to potential energy. That is, (1/2)mv^2 = mgh, where h is the vertical displacement. Solving for v gives v = sqrt(2gh). Draw a right triangle and let the hypotenuse length be 5 m. Then h = 5 sin (theta), where theta = arcsin(3/5) making h = 3m. Therefore,

v = sqrt(2gh) = sqrt (6g). I leave you and your calculator to finish it up.

Regards,

Rich B.

10. Sorry Ron. Apparently we constructed our replies simultaneously and your superior typing skills won out.

...Rich

11. Originally Posted by Rich B.
Sorry Ron. Apparently we constructed our replies simultaneously and your superior typing skills won out.

...Rich
Not to worry.

(But as an aside in a long-ish thread like this you should quote what you

RonL

12. Originally Posted by CaptainBlack
Not to worry.

(But as an aside in a long-ish thread like this you should quote what you

RonL
...like this?

13. Originally Posted by Rich B.
...like this?
Excelent -why isn't there an emoticom of Mr Burns when you need one

RonL

14. Originally Posted by CaptainBlack
Excelent -why isn't there an emoticom of Mr Burns when you need one

RonL
They got "Simpson's" in England?

15. Originally Posted by ThePerfectHacker
They got "Simpson's" in England?
Yes, why not?

RonL

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