Author Topic: Dynamics (Read 14544 times) Tweet Share

/0 · « **Reply #15 on:** August 12, 2009, 10:12:02 pm »

Quote from: d0minicz on August 12, 2009, 10:10:17 pm

ah thanks man
So is kg wt the same as Newtons ?

Yep. I believe it is defined as the weight of 1kg, which would be 1N.
NO IT ISNT

kamil9876 · « **Reply #16 on:** August 12, 2009, 10:14:20 pm »

nah. 1kg weight is 9.8N.

It's like when someone says "I weigh 100kg" they really mean the weight force is equal to that of a 100kg mass in earth's gravitational field, hence really 980N. I think.

/0 · « **Reply #17 on:** August 12, 2009, 10:20:22 pm »

lol oops

d0minicz · « **Reply #18 on:** August 12, 2009, 10:24:37 pm »

lol oh
wait sooo in the book they have the formula : R - mg = ma , and they say R is in newtons
why were you able to use 2.5 kg wt in that formula and get the correct answer lol confused now

d0minicz · « **Reply #19 on:** August 13, 2009, 05:12:57 pm »

The engine of a train of mass 200 tonnes exerts a force of 8000 kg wt, and the total air and rail resistance is 20 kg wt/tonne. How long will it take the train on level ground to acquire a speed of 30 km/h from rest?

thanks.

Damo17 · « **Reply #20 on:** August 13, 2009, 05:44:11 pm »

Quote from: d0minicz on August 13, 2009, 05:12:57 pm

The engine of a train of mass 200 tonnes exerts a force of 8000 kg wt, and the total air and rail resistance is 20 kg wt/tonne. How long will it take the train on level ground to acquire a speed of 30 km/h from rest?

thanks.

$F_{net}=(8000 \times 9.8) - (20 \times 9.8 \times 200) = (200 \times 10^3)a$

$\therefore a= 0.196$

$u=0$ , $v= \frac{30}{3.6}$ , $a=0.196$ , $t=?$

$v=u+at$

$\frac{30}{3.6}=0.196t$

$\therefore t=42.517s$

Just tell me if you don't understand why I did something and I will explain it.

d0minicz · « **Reply #21 on:** August 13, 2009, 06:12:51 pm »

ah thanks

i dont really understand the first two steps ; like where did the a come from in the first line and how did you solve for a in the second step?
sorry just trying to get used to this topic
appreciate it

Damo17 · « **Reply #22 on:** August 13, 2009, 06:25:15 pm »

Quote from: d0minicz on August 13, 2009, 06:12:51 pm

ah thanks

i dont really understand the first two steps ; like where did the a come from in the first line and how did you solve for a in the second step?
sorry just trying to get used to this topic
appreciate it

$a=acceleration$

Firstly I found $F_{net}$ , since the engine exerts a driving force of $8000 kg \ wt$ , which is $(8000 \times 9.8) N$ and the resistive force is $20kg \ wt/tonne$ , which is $(20 \times 9.8)N/tonne$ so because it has a mass of 200tonnes, Total resisitive Force is $(20 \times 9.8 \times 200)N$

$F_{net}= F_{driving} - F_{resistive} = 78400-39200=39200N$

Now we know $F_{net}= ma = (200 \times 10^3) \times a = \frac{39200}{200 \times 10^3} = 0.196ms^{-2}$

we use $200 \times 10^3$ because we want mass in $kg$ , not $tonnes$ .

d0minicz · « **Reply #23 on:** August 15, 2009, 06:55:40 pm »

One man can push a wardrobe of mass 250kg with an acceleration of magnitude 0.15 m/s^2 . With the help from another man pushing it just as hard (ie. with the same force), the wardrobe accelerates at 0.4 m/s^2. How hard is each man pushing and what is the resistance to sliding?

thanks...

kamil9876 · « **Reply #24 on:** August 15, 2009, 07:28:15 pm »

in generah, if $n$ men push the thing, each with force $F$ , and a resistance force of $R$ then:

$F_{net}=nF-R$

for one man this is:
$<br />F_{net}=F-R$

for two it is:

$F_{net}=2F-R$

Subbing in $F_{net}=ma$ :
$250*0.15=F-R$
$250*0.4=2F-R$

Now solve simultaenously.

bigtick · « **Reply #25 on:** August 15, 2009, 08:56:37 pm »

A similar question from another forum:
A person (65 kg) pushes horizontally a sliding crate (80 kg) on a level floor. The coefficient of friction between the floor and the person or crate is 0.2. Find the maximum magnitude of the acceleration of the crate. g = 9.8 N/kg. (Ans. 2.0 m/s2)

Damo17 · « **Reply #26 on:** August 15, 2009, 09:06:59 pm »

Quote from: bigtick on August 15, 2009, 08:56:37 pm

A similar question from another forum:
A person (65 kg) pushes horizontally a sliding crate (80 kg) on a level floor. The coefficient of friction between the floor and the person or crate is 0.2. Find the maximum magnitude of the acceleration of the crate. g = 9.8 N/kg. (Ans. 2.0 m/s2)

I think:

$F_{friction}= (65 \times 9.8 \times 0.2)+(80 \times 9.8 \times 0.2)=284.2$

since the crate is sliding, $F_{net}=284.2$

$F_{net}=ma$
$284.2=(80+65)a$
$a=1.96ms^{-2}$

bigtick · « **Reply #27 on:** August 15, 2009, 09:28:44 pm »

The second part of the question:

If the coefficient of friction between the floor and the person is increased to 0.3, find the maximum rate of increase in the speed of the crate.

The answer to the first part was changed to 3.6 m/s2.

d0minicz · « **Reply #28 on:** August 16, 2009, 09:07:47 pm »

thanks

~~1. What force is necessary to accelerate a train of mass 200 tonnes at 0.2 m/s^2 against a resistance of 20 000N?~~ What will be the acceleration if the train free-wheels against the same resistance?

2. A puck of mass 0.1 kg is sliding in straight line on an ice-rink. The coefficient of friction between the puck and the use is 0.025.
~~a) Find the resistive force owing to friction. Ans: 0.0245N~~
b) Find the speed of the puck after 20 seconds if its initial speed is 10 m/s.

thx again

d0minicz · « **Reply #29 on:** August 20, 2009, 05:53:51 pm »

1. A body mass of M kg is pulled along a rough horizontal plane (coefficient of friction $\mu$ ) by a constant force of F newtons, at an inclination of $\theta$ . Find the acceleration of the body if:

~~a) $\theta$ is upwards from the horizontal~~

b) $\theta$ is downwards from the horizontal.

2. A particle of mass 5kg is being pulled up a slope inclined at $30^o$ to the horizontal. The pulling force, F newtons, acts parallel to the slope, as does the resistance with a magnitude of one-fifth of the magnitude of the normal reaction.

~~a) Find the value of F, such that the acceleration is $1.5 m/s^2$ up the slope.~~

b) Also find the magnitude of the acceleration if this pulling force now acts at an angle of $30^o$ to the slope (ie. at $60^o$ to the horizontal).

thanksssss appreciate it

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