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May 17, 2025, 03:21:05 pm
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Author Topic: Dynamics  (Read 13256 times)  Share 

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d0minicz

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Re: Dynamics
« Reply #45 on: August 23, 2009, 08:31:21 pm »
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haha thanks guys
hey for finding velocity, do i use the t > 5 function only?
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d0minicz

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Re: Dynamics
« Reply #46 on: August 23, 2009, 09:32:53 pm »
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Another one...

A particle of mass m kg at rest on a horizontal plane is acted on by a constant horizontal force b N. The total resistance to motion is cv N where v m/s is the velocity and c is a constant value. Find the velocity at time t seconds and the terminal velocity.

thanks...
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/0

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Re: Dynamics
« Reply #47 on: August 23, 2009, 09:51:10 pm »
0














,



, where

At t = 0, v = 0



As ,

hope i haven't made any misteaks
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d0minicz

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Re: Dynamics
« Reply #48 on: August 24, 2009, 04:19:56 pm »
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Another one of those:

A body of mass m is projected vertically upwards with speed u. Air resistance is equal to k times the square of the speed where k is constant. Find the maximum height reached and the speed when next at the point of projection.

thx alot lol
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/0

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Re: Dynamics
« Reply #49 on: August 24, 2009, 04:43:11 pm »
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Lol I remember this from earlier (midway down the page): http://vcenotes.com/forum/index.php/topic,9192.165.html ;p
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d0minicz

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Re: Dynamics
« Reply #50 on: August 24, 2009, 10:59:21 pm »
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lol thanks nice one

Sorry another, i just cant get the diagram right ...

A particle of mass 5kg hangs from a fixed point O by a light inextensible string. it is pulled aside by a force of P N that makes an angle of with the downward vertical and rests in equilibrium with the string inclined at to the vertical. Find P.

thanks ....
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/0

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Re: Dynamics
« Reply #51 on: August 24, 2009, 11:09:22 pm »
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I think this must be how the diagram is set up. Force P must be to the right so it can balance the tension and keep the particle in equilibrium.
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d0minicz

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Re: Dynamics
« Reply #52 on: August 26, 2009, 09:11:00 pm »
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ah thanks =]

Another...

A particle of mass 5kg is suspended by two strings of lengths 5cm and 12cm respectively, attached at two points at the same horizontal level and 13cm apart. Find the tension in the shorter string...

thanks again
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kamil9876

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Re: Dynamics
« Reply #53 on: August 26, 2009, 10:03:07 pm »
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tricky one:

notice that which means that the angle between the two ropes is 90degrees since the converse of Pythagoras' theorem is true.

Using knowledge of this you can show that the alternate angles are equal and so you can use trigonometry to play around with those angles and decompose the two tensions in components and add the the three forces(two tensions and gravity) and get some simultaenous equations.
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d0minicz

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Re: Dynamics
« Reply #54 on: August 27, 2009, 05:28:48 pm »
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1. A particle of mass 1.2kg rests on a rough horizontal table. The coefficient of friction between the particle and the table is 0.2. A force P N is applied to the particle along a string attached to it. The string is inclined at an angle of  to the table. Find the friction force, correct to two decimal places, if:
a) P = 2
b) P = 3

2. A particle weighing 3 kg wt rests in limiting equilibrium on a rough plane inclined at an angle of to the horizontal. Find:
a) the coefficient of friction between the particle and the plane. Ans: 0.47
b) the least force which when applied up the plane will cause the particle to move

3. A mass of 24 kg is on the point of motion down a rough inclined plane when supported by a force of 100 N parallel to the plane. If the magnitude of the force is increased to 120 N, the mass is on the point of moving up the plane. Find:
a) the friction force
b) the inclination of the plane to the horizontal
c) the coefficient of friction

sorry for the heap of q's ... having trouble
please help =]

thanks ..
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TrueTears

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Re: Dynamics
« Reply #55 on: August 27, 2009, 06:06:55 pm »
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1. a) Consider the forward force. This will be equal to 2cos40.

The retarding force is denoted by



so

Therefore resultant F = ma

Now notice the F_f is constant, only acceleration is affected. Thus F_f =0.2 *(2sin40 - 1.2 *9.8)
« Last Edit: August 27, 2009, 06:13:41 pm by TrueTears »
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d0minicz

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Re: Dynamics
« Reply #56 on: August 27, 2009, 09:53:56 pm »
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ANS for them just incase u guys wanted to check;
1.a) 1.53N
b) 1.97N

2.b) 24.86N

3.a)10N
b) 27.88 degrees
c) 0.05

anyone got any ideas... please desperate lollllll hehehe
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kamil9876

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Re: Dynamics
« Reply #57 on: August 27, 2009, 10:27:43 pm »
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2b.)

The greatest the friction can be is .

If we consider up the slope as the positive direction, we essentially want a positive acceleration:



(since )
(Friction is acting down the plane when particle is moving up)



Hence the minimum force is the LHS of the inequality.



isn't TT's ans to q1 correct btw?
« Last Edit: August 27, 2009, 11:20:39 pm by kamil9876 »
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ed_saifa

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Re: Dynamics
« Reply #58 on: August 27, 2009, 10:34:42 pm »
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Qn 3
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d0minicz

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Re: Dynamics
« Reply #59 on: August 27, 2009, 10:55:53 pm »
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ahh thanks guys ...

Kamil regarding question 2, im unsure of what to sub into that equation you provided me with
so i have mu = 0.47 , R = 3cos(25) , but what to sub for the is what im unsure about because we're given 3 kg wt...

oh for TT's solution, im confused of what to put into the calc. if i put his final line into the calc i get a different answer.
maybe im missing something =/

nevertheless thanks alot for that
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