Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    http://www.mei.org.uk/files/papers/M1_Paper_Jan13.pdf
    For the above above paper question 3iii, why couldn't you just resolve perpendicular to motion and say that there is a resultant force because Sam's and Tom's components aren't the same at right angles to motion. (the mark scheme is at the bottom of the pdf)
    Also for question 7vi, i don't get whats going on

    Thank you
    Online

    22
    ReputationRep:
    (Original post by runny4)
    http://www.mei.org.uk/files/papers/M1_Paper_Jan13.pdf
    For the above above paper question 3iii, why couldn't you just resolve perpendicular to motion and say that there is a resultant force because Sam's and Tom's components aren't the same at right angles to motion. (the mark scheme is at the bottom of the pdf)
    Uh, you can? I think - it's a fine argument. There's a component of force acting at right-angles to the motion, hence the resultant force will not be in the direction of the motion.


    Also for question 7vi, i don't get whats going on

    Thank you
    The part of the string BC is 2.3 metres which is greater than 2 metres, i.e: it is "too long" and the piece BC will be effectively useless. Imagine holding a block using a chain/string that is 2 metres long when the block is situated one metre away from you, the chain/string will be pretty much useless, it'll be just limp (slack) and doing nothing. That's the case here: the piece AB will have to handle the entirety of the weight (50N) and you'll find that the angle CAB is 90 degrees, so it's basically just resolve upwards/downwards to get AC = 50 N and BC = 0 / useless.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Uh, you can? I think - it's a fine argument. There's a component of force acting at right-angles to the motion, hence the resultant force will not be in the direction of the motion.




    The part of the string BC is 2.3 metres which is greater than 2 metres, i.e: it is "too long" and the piece BC will be effectively useless. Imagine holding a block using a chain/string that is 2 metres long when the block is situated one metre away from you, the chain/string will be pretty much useless, it'll be just limp (slack) and doing nothing. That's the case here: the piece AB will have to handle the entirety of the weight (50N) and you'll find that the angle CAB is 90 degrees, so it's basically just resolve upwards/downwards to get AC = 50 N and BC = 0 / useless.
    Just for the 2nd question if you use the cosine rule to find angle CAB, you get
    (4+0.25-2.3^2)/(2*2*0.5)= cos(CAB) but this does not give you CAB=90
    Online

    22
    ReputationRep:
    (Original post by runny4)
    Just for the 2nd question if you use the cosine rule to find angle CAB, you get
    (4+0.25-2.3^2)/(2*2*0.5)= cos(CAB) but this does not give you CAB=90
    Right, but if you're using the cosine rule like that, then you're assuming BC is taut when it isn't.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Right, but if you're using the cosine rule like that, then you're assuming BC is taut when it isn't.
    Then how would you deduce that Angle CAB is 90 degrees?
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by runny4)
    http://www.mei.org.uk/files/papers/M1_Paper_Jan13.pdf
    For the above above paper question 3iii, why couldn't you just resolve perpendicular to motion and say that there is a resultant force because Sam's and Tom's components aren't the same at right angles to motion. (the mark scheme is at the bottom of the pdf)
    IMO, saying there is a resultant force perpendicular to the acceleration isn't sufficient. There must be something counteracting that resultant force for the car to accelerate as it does. I.e. a resistive force perpendicular to the motion.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    IMO, saying there is a resultant force perpendicular to the acceleration isn't sufficient. There must be something counteracting that resultant force for the car to accelerate as it does. I.e. a resistive force perpendicular to the motion.
    but its asking why the resultant of the forces exerted by Sam and Tom is not in the direction of the car’s acceleration not why the car travels in a straight line
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by runny4)
    but its asking why the resultant of the forces exerted by Sam and Tom is not in the direction of the car’s acceleration not why the car travels in a straight line
    it's not about why the car travels in a straight line.

    Your answer is only showing that the sum of the perpendicular components of the forces (of the two people) is non-zero. This does not explain why the car moves in the direction it does.

    The car will accelerate in the direction of the sum of the forces acting on it - always. The question is asking why is this direction different to the resultant of the forces exerted by the two people.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    it's not about why the car travels in a straight line.

    Your answer is only showing that the sum of the perpendicular components of the forces (of the two people) is non-zero. This does not explain why the car moves in the direction it does.

    The car will accelerate in the direction of the sum of the forces acting on it - always. The question is asking why is this direction different to the resultant of the forces exerted by the two people.
    ok thanks
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Have you ever participated in a Secret Santa?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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