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doesn't part C of this question contradict what was asked for in part b? Watch

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    Question is attached. I accept part b has nothing wrong with it (solution attached) but part c clearly shows (part c solution is given as part of partb solution) that friction = 0.3 R = 0.3x(Mg+W) and so the mass does affect friction and so friction is not independent of the mass as stated in part b. I am totally split two ways by this question. Anyone who can help?
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    (Original post by thebrahmabull)
    Question is attached. I accept part b has nothing wrong with it (solution attached) but part c clearly shows (part c solution is given as part of partb solution) that friction = 0.3 R = 0.3x(Mg+W) and so the mass does affect friction and so friction is not independent of the mass as stated in part b. I am totally split two ways by this question. Anyone who can help?
    I can't see the solution very clearly (maybe I'm getting old), but it looks like it says something like F \leq 0.3R=0.3(Mg+W) rather than F = 0.3R?
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    (Original post by Implication)
    I can't see the solution very clearly (maybe I'm getting old), but it looks like it says something like F \leq 0.3R=0.3(Mg+W) rather than F = 0.3R?
    Yes. But mew x R is friction by mathematical definition right?
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    (Original post by thebrahmabull)
    Yes. But mew x R is what friction is right?
    \mu R is the maximum force due to friction I believe. The actual force, denoted by F in this question, will be less than (or equal to) that.

    I could be wrong though. It's been 4 years since I did A-level maths so I'm a bit fuzzy on this stuff
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    No. In B you were finding a formula for the friction force. In C you are applying a constraint on the friction.
    (Original post by thebrahmabull)
    Yes. But mew x R is friction by mathematical definition right?
    No. Friction defined something like the force, parallel to the surface, that an object experiences due to interactions with the surface.

    The case F=\mu R only applies when the object is moving or in limiting equilibrium.
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    (Original post by Implication)
    \mu R is the maximum force due to friction I believe. The actual force, denoted by F in this question, will be less than (or equal to) that.

    I could be wrong though. It's been 4 years since I did A-level maths so I'm a bit fuzzy on this stuff
    (Original post by morgan8002)
    No. In B you were finding a formula for the friction force. In C you are applying a constraint on the friction.


    No. Friction defined something like the force, parallel to the surface, that an object experiences due to interactions with the surface.

    The case F=\mu R only applies when the object is moving or in limiting equilibrium.
    So should I always visualise that mew x R is something different from friction, e.g.its a horizontal force, but not friction?
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    (Original post by thebrahmabull)
    So should I always visualise that mew x R is something different from friction, e.g.its a horizontal force, but not friction?
    No. Visualise it as the maximum possible value of friction.
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    (Original post by thebrahmabull)
    So should I always visualise that mew x R is something different from friction, e.g.its a horizontal force, but not friction?
    \mu R is the friction, but only in the limiting case i.e. when the object is moving or on the point of moving. Think of it as the maximum friction so F_{\mathrm{max}}=\mu R. The more you try to move something, the stronger the frictional force resisting the motion gets - until it reaches its maximum value, when the object begins to move. If the object isn't moving and you aren't told it's in limiting equilibrium, you have to find the friction another way.
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    Thanks guys!
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    (Original post by morgan8002)
    No. In B you were finding a formula for the friction force. In C you are applying a constraint on the friction.


    No. Friction defined something like the force, parallel to the surface, that an object experiences due to interactions with the surface.

    The case F=\mu R only applies when the object is moving or in limiting equilibrium.
    Interesting question, I have read the thread but part B confuses me. How can friction exerted by ladder be independent of the mass of the brick? Since the ladder is not in equilibrium, it is now moving, so maximum friction must be acting on it... Mu x R value gets greater as mass of the brick increases till it becomes (3w/8) right ?
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    (Original post by chhhhelsie)
    Interesting question, I have read the thread but part B confuses me. How can friction exerted by ladder be independent of the mass of the brick? Since the ladder is not in equilibrium, it is now moving, so maximum friction must be acting on it... Mu x R value gets greater as mass of the brick increases till it becomes (3w/8) right ?
    We don't know that it's not in equilibrium though. Friction could be less than \mu R.
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    (Original post by morgan8002)
    We don't know that it's not in equilibrium though. Friction could be less than \mu R.
    Ohh right
 
 
 
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