Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
    Could someone help me out with the second part of this question?
    Attached Images
     
    Offline

    19
    ReputationRep:
    (Original post by Firefox23)
    Could someone help me out with the second part of this question?
    you need to attach it...
    Offline

    22
    ReputationRep:
    What ave you tried so far?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    What ave you tried so far?
    I initially tried to find the acceleration using F=man with F as the weight down the plane - the friction up the plane and used s=ut +1/2at^2 with s as 5a and u as 0
    Offline

    2
    ReputationRep:
    draw a clear diagram first then label all of the forces along each plane and you should be able to do this rather easily.
    Offline

    22
    ReputationRep:
    (Original post by Firefox23)
    I initially tried to find the acceleration using F=man with F as the weight down the plane - the friction up the plane and used s=ut +1/2at^2 with s as 5a and u as 0
    Okay, so we're taking down the slop as positive then - this means we have

    s = 5a, u=0, a = a, t=T so we need only find acceleration then.

    Let's see - we resolve down the plane to get mg \sin \theta - \mu R where we can get R easily, simply resolve perpendicular to the plane to get that R = mg \cos \theta

    So our force down the plane is mg \sin \theta - \frac{1}{4} \mg \cos \theta where \theta = \arctan \frac{3}{4}.

    Hence you get (after dividing by m) a = \frac{2g}{5}.

    Using the suvat equation yields

    5a = \frac{1}{2} \cdot \frac{2g}{5} \cdot T^2 \implies T = 5\sqrt{\frac{a}{g}}

    That's the time down the slope, they are asking for the total time - so you need to add the time to go up the slope as well.

    That's a basic suvat solve for let's call it t_{up}, which you should get as
    Spoiler:
    Show
    t_{up} = \frac{5}{2} \sqrt{\frac{2a}{g}}


    Then you add and pull out a common factor.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Okay, so we're taking down the slop as positive then - this means we have

    s = 5a, u=0, a = a, t=T so we need only find acceleration then.

    Let's see - we resolve down the plane to get mg \sin \theta - \mu R where we can get R easily, simply resolve perpendicular to the plane to get that R = mg \cos \theta

    So our force down the plane is mg \sin \theta - \frac{1}{4} \mg \cos \theta where \theta = \arctan \frac{3}{4}.

    Hence you get (after dividing by m) a = \frac{2g}{5}.

    Using the suvat equation yields

    5a = \frac{1}{2} \cdot \frac{2g}{5} \cdot T^2 \implies T = 5\sqrt{\frac{a}{g}}

    That's the time down the slope, they are asking for the total time - so you need to add the time to go up the slope as well.

    That's a basic suvat solve for let's call it t_{up}, which you should get as
    Spoiler:
    Show
    t_{up} = \frac{5}{2} \sqrt{\frac{2a}{g}}

    Then you add and pull out a common factor.
    Thanks so much I forgot to add the time up the slope to get the total time. Any chance you could help me with this question? I'm not sure where to start.
    Attached Images
     
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Firefox23)
    Any chance you could help me with this question? I'm not sure where to start.
    Start at A.

    Since you have symmetry, what's the vertical reaction at A?

    As A and E are as far apart as possible what's the frictional force going to be at A?

    What's the vertical component in AB?

    You now have sufficient information to work out the angle of AB, and hence the horizontal displacment of B.

    Now go on to consider B and BC.
    PS Not worked that bit through.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    Start at A.

    Since you have symmetry, what's the vertical reaction at A?

    As A and E are as far apart as possible what's the frictional force going to be at A?

    What's the vertical component in AB?

    You now have sufficient information to work out the angle of AB, and hence the horizontal displacment of B.

    Now go on to consider B and BC.
    PS Not worked that bit through.
    The tension in each section of the string will be different right? Also what do you mean by the vertical component in AB? This is what I tried . . .
    Attached Images
     
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Firefox23)
    The tension in each section of the string will be different right?
    The tension in the upper strings will be different to that in the lower strings.

    Also what do you mean by the vertical component in AB? This is what I tried . . .
    The tension in AB can be split into vertical and horizontal components.


    You can write down the vertical reaction at A straight off. What's the total weight? Then by symmetry vertical reaction at A is ...?

    Then as per my previous questions.

    Note: There is no T_1 force, only friction and the horizontal component in AB.
    • Thread Starter
    Offline

    2
    ReputationRep:
    Name:  image.jpg
Views: 36
Size:  495.1 KB
    (Original post by ghostwalker)
    The tension in the upper strings will be different to that in the lower strings.



    The tension in AB can be split into vertical and horizontal components.


    You can write down the vertical reaction at A straight off. What's the total weight? Then by symmetry vertical reaction at A is ...?

    Then as per my previous questions.

    Note: There is no T_1 force, only friction and the horizontal component in AB.
    This is what I've tried but I still seem to have too many unknowns . . .
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Firefox23)
    Name:  image.jpg
Views: 36
Size:  495.1 KB
    This is what I've tried
    Doesn't seem that way from your working - unless you've done something since my last post, but haven't posted it.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    Doesn't seem that way from your working - unless you've done something since my last post, but haven't posted it.
    So my working's correct?
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Firefox23)
    So my working's correct?
    Sorry, didn't see your new working at the top of the post - will have a look now.
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Firefox23)
    So my working's correct?
    You have ignored what I suggested.

    The total weight of the mass is 8mg.
    The vertical reactions at A and E must add to 8mg.

    By symmetry the vertical reaction at A will be half that, i.e. 4mg.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    You have ignored what I suggested.

    The total weight of the mass is 8mg.
    The vertical reactions at A and E must add to 8mg.

    By symmetry the vertical reaction at A will be half that, i.e. 4mg.
    Does the tension in AB not affect the vertical reaction? Sorry I didn't misread the question and didn't notice that the A and E had different masses to B,C and D
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Firefox23)
    Does the tension in AB not affect the vertical reaction? Sorry I didn't misread the question and didn't notice that the A and E had different masses to B,C and D
    The tension in AB does effect the vertical reaction. The vertical reaction at A, 4mg, arises as a result of the weight of A, mg, and the vertical component of the tension in AB.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    The tension in AB does effect the vertical reaction. The vertical reaction at A, 4mg, arises as a result of the weight of A, mg, and the vertical component of the tension in AB.
    I think I understand R=4mg and it also = mg +tcostheta
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Firefox23)
    I think I understand R=4mg and it also = mg +tcostheta
    Yep.

    In the first case you look at the structure as a whole, and in the second you just focus on the details at A.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    Yep.

    In the first case you look at the structure as a whole, and in the second you just focus on the details at A.
    Thanks I've worked it through now, any chance you could help me out with this question if you have time? I've tried to do velocity triangles and I've tried i and j notation but I can't seem to get the right answer. Its question 14.Name:  FullSizeRender-9.jpg
Views: 84
Size:  70.6 KB
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: September 24, 2015

University open days

  1. University of Bradford
    University-wide Postgraduate
    Wed, 25 Jul '18
  2. University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Wed, 25 Jul '18
  3. Bournemouth University
    Clearing Campus Visit Undergraduate
    Wed, 1 Aug '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.