My Math Forum Using Newton's laws etc. pulley type.

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 March 22nd, 2016, 07:04 AM #1 Newbie   Joined: Mar 2016 From: Physics land Posts: 7 Thanks: 0 Using Newton's laws etc. pulley type. Consider a situation where there is a particle1 which has mass 6kg. The particle is momentarily at rest at an angle of 42 degrees above the horizontal. The particle is then connected by an inextensible, light string to another particle 2 with mass of 3.5kg. Now, this string then goes over a pulley where friction is negligible. Refer to the diagram below. The coefficient of friction (kinematical and statics) relating the question which is contained in between the particle m1 and the slope's surface have values of statics friction = 9/20 and kinematical friction= 7/20. Additional information provided: m1 = 6kg m2 = 3.5kg theta = 42 degrees With all the information provided, deduce whether particle1 will move and if so state the direction, or remain at its current position. The slope angle theta is again 42 degrees, what would be the mass of particle 3 (a different particle with different mass), which is connected to the string which will cause the particle1 of mass 6kg to accelerate in an upwards direction where the acceleration is 3.81m/s^2? In your answer you must clearly state the equations(SUVAT) that you have used for each particle. Thanks if you read all this!
 March 22nd, 2016, 08:08 AM #2 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 So what do you think? Have you attempted the first part yet? Hint: To solve the first part you need to obtain an inequality. This will tell you whether the mass on the slope is in equilibrium or not. Thanks from devour19 Last edited by studiot; March 22nd, 2016 at 08:10 AM.
March 22nd, 2016, 08:42 AM   #3
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 Originally Posted by studiot So what do you think? Have you attempted the first part yet? Hint: To solve the first part you need to obtain an inequality. This will tell you whether the mass on the slope is in equilibrium or not.

I am still unsure what inequality to devise. Please can you advise me further in what I should use to construct it.

Usually I resolve in the x and y directions..

March 22nd, 2016, 09:37 AM   #4
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 Usually I resolve in the x and y directions..
Then what do you do with the resolutes?

What is the condition (involving friction) for slipping/not slipping between any contact surfaces?

Is this not the inequality you need?

 March 22nd, 2016, 09:47 AM #5 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 disregarding friction for a moment ... if $m_2g > m_1g\sin{\theta}$, then $m_1$ will slide up the incline now, throw friction in ... if $f_s$ is large enough,then equilibrium will be achieved and $m_2g = m_1g\sin{\theta} + f_s$ $m_2g - m_1g\sin{\theta} = f_s$ recall $f_s \le \mu_s \cdot m_1g\cos{\theta}$ ... $m_2g - m_1g\sin{\theta} = f_s \le \mu_s \cdot m_1g\cos{\theta}$ $m_2g - m_1g\sin{\theta} \le \mu_s \cdot m_1g\cos{\theta}$ $m_2 - m_1\sin{\theta} \le \mu_s \cdot m_1\cos{\theta}$ now consider the other possibility ... if $m_2g < m_1g\sin{\theta}$, then $m_1$ will slide down the incline and the force of static friction will be directed up the incline. Of course, if $f_{s \, max}$ is not great enough, then you will have a dynamic situation where $m_1$ slides up or down the incline depending on the relative magnitudes of $m_2g$ and $m_1g\sin{\theta}$ Thanks from devour19
March 22nd, 2016, 01:21 PM   #6
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 Originally Posted by studiot Then what do you do with the resolutes? What is the condition (involving friction) for slipping/not slipping between any contact surfaces? Is this not the inequality you need?
I am an A Level Student.

What is resolute?

F=uR, newton's coefficient of friction right? But I have no idea what kinematic friction is I only know of regular basic? statics work.

Thanks.

March 22nd, 2016, 01:24 PM   #7
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 Originally Posted by skeeter disregarding friction for a moment ... if $m_2g > m_1g\sin{\theta}$, then $m_1$ will slide up the incline now, throw friction in ... if $f_s$ is large enough,then equilibrium will be achieved and $m_2g = m_1g\sin{\theta} + f_s$ $m_2g - m_1g\sin{\theta} = f_s$ recall $f_s \le \mu_s \cdot m_1g\cos{\theta}$ ... $m_2g - m_1g\sin{\theta} = f_s \le \mu_s \cdot m_1g\cos{\theta}$ $m_2g - m_1g\sin{\theta} \le \mu_s \cdot m_1g\cos{\theta}$ $m_2 - m_1\sin{\theta} \le \mu_s \cdot m_1\cos{\theta}$ now consider the other possibility ... if $m_2g < m_1g\sin{\theta}$, then $m_1$ will slide down the incline and the force of static friction will be directed up the incline. Of course, if $f_{s \, max}$ is not great enough, then you will have a dynamic situation where $m_1$ slides up or down the incline depending on the relative magnitudes of $m_2g$ and $m_1g\sin{\theta}$
Thank you. This is helpful I will see where this leads me.

March 22nd, 2016, 02:16 PM   #8
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 What is resolute?
resolute on its own is an adjective which means determined or steadfast.
the resolute is a noun which is the result of resolving vectors (forces etc)

Since skeeter has decided to do you homework for you I will not interfere.

However I will explain about friction.

The force of friction is not constant.
Before an object moves the frictional force is only exactly enough to oppose the disturbing or driving force (say D)
So it rises from zero when there is no driving force up to a maximum value when object is just about to move.
At this moment the value equals the coefficient of friction times the normal reaction between the surfaces F = uR.
As the driving force increases from zero to D, the friction increases from zero to F = uR which is always equal to the driving force so the body is alway in equilibrium and the rules of statics apply.
u is called the coefficient of static friction.

Once the body starts to move the coefficient of friction drops slightly and remains constant.
This lower coefficient of friction is called the coefficient of dynamic friction.
Because this coefficient is constant, the frictional force no longer varies as the driving force increases but also remains constant.
Thus the net force on the body (equal to D - F) increases as D increases as D increases.
In these circumstances Newtons laws of motion apply and the acceleration = mass times (D-F)

March 22nd, 2016, 03:22 PM   #9
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 Originally Posted by skeeter disregarding friction for a moment ... if $m_2g > m_1g\sin{\theta}$, then $m_1$ will slide up the incline now, throw friction in ... if $f_s$ is large enough,then equilibrium will be achieved and $m_2g = m_1g\sin{\theta} + f_s$ $m_2g - m_1g\sin{\theta} = f_s$ recall $f_s \le \mu_s \cdot m_1g\cos{\theta}$ ... $m_2g - m_1g\sin{\theta} = f_s \le \mu_s \cdot m_1g\cos{\theta}$ $m_2g - m_1g\sin{\theta} \le \mu_s \cdot m_1g\cos{\theta}$ $m_2 - m_1\sin{\theta} \le \mu_s \cdot m_1\cos{\theta}$ now consider the other possibility ... if $m_2g < m_1g\sin{\theta}$, then $m_1$ will slide down the incline and the force of static friction will be directed up the incline. Of course, if $f_{s \, max}$ is not great enough, then you will have a dynamic situation where $m_1$ slides up or down the incline depending on the relative magnitudes of $m_2g$ and $m_1g\sin{\theta}$
Thank you I have calculated from your help that I have the m1 will slide down the slope, thank you for the clarity in your calculation.

Finally I am unsure for the second part - would I rearrange the equation to make the mass of the second particle the subject of the equation? And from the conditions you have said.

Thank you.

March 22nd, 2016, 03:34 PM   #10
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Quote:
 Originally Posted by skeeter disregarding friction for a moment ... if $m_2g > m_1g\sin{\theta}$, then $m_1$ will slide up the incline now, throw friction in ... if $f_s$ is large enough,then equilibrium will be achieved and $m_2g = m_1g\sin{\theta} + f_s$ $m_2g - m_1g\sin{\theta} = f_s$ recall $f_s \le \mu_s \cdot m_1g\cos{\theta}$ ... $m_2g - m_1g\sin{\theta} = f_s \le \mu_s \cdot m_1g\cos{\theta}$ $m_2g - m_1g\sin{\theta} \le \mu_s \cdot m_1g\cos{\theta}$ $m_2 - m_1\sin{\theta} \le \mu_s \cdot m_1\cos{\theta}$ now consider the other possibility ... if $m_2g < m_1g\sin{\theta}$, then $m_1$ will slide down the incline and the force of static friction will be directed up the incline. Of course, if $f_{s \, max}$ is not great enough, then you will have a dynamic situation where $m_1$ slides up or down the incline depending on the relative magnitudes of $m_2g$ and $m_1g\sin{\theta}$
finally I am unsure how to link the equations to suvat, I gather I would have a and v obviously, since I won't be able to have s or t it means I must need u which I don't know how to figure it.

Thank you sir, this question has been bothering me all day.

 Tags laws, newton, pulley, type

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