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  1. #1
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    help understanding answer

    https://www.examinations.ie/tmp/1495...9333.pdf#page=

    could someone explain question 7b to me. on the link you will find the answer to 7b but i do not understand the answer.
    i do not understand why they write R2<R1 and i don't see how this implies F2=uR2 and i also don't understand how this gives the least value of u necessary for equilibrium
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  2. #2
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    Re: help understanding answer

    You seem to have posted a link to an information page about the Irish "State Examination Council" rather than a single problem or a single test. How are we to find "problem 7b" on that?
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    Re: help understanding answer

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    Re: help understanding answer

    any help please?
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    Re: help understanding answer

    Start by separating the forces acting on rods ab and bc at points a and b. The weight of each is directed into the ground at angle \pi/2- \theta (since they are of the same length, angle 2\theta is divided equally). Separate that force into horizontal and vertical components. In order that there be no slip, the friction force, \mu times the vertical components, must be at least equal to the horizontal force.
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    Re: help understanding answer

    Quote Originally Posted by markosheehan View Post
    https://www.examinations.ie/tmp/1495...9333.pdf#page=

    i do not understand why they write R2<R1 and i don't see how this implies F2=uR2 and i also don't understand how this gives the least value of u necessary for equilibrium
    I do not agree with that "equation" ... it should be an inequality since $f_s \le \mu \cdot N$

    I assume you're ok with how $R_1 = \dfrac{7W}{4}$ and $R_2 = \dfrac{5W}{4}$ were calculated.

    I also assume you're ok with how $F_1 = F_2 = \dfrac{3W}{4} \cdot \tan{\theta}$ was calculated, also.

    now ...

    $F_1 \le \mu \cdot R_1 \implies \dfrac{3W}{4} \tan{\theta} \le \mu \cdot \dfrac{7W}{4} \implies \mu \ge \dfrac{3}{7} \cdot \tan{\theta}$

    $F_2 \le \mu \cdot R_2 \implies \dfrac{3W}{4} \tan{\theta} \le \mu \cdot \dfrac{5W}{4} \implies \mu \ge \dfrac{3}{5} \cdot \tan{\theta}$

    the more restrictive inequality of the two above gives the minimum value for the coefficient of friction, $\mu \ge \dfrac{3}{5} \cdot \tan{\theta}$
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    Re: help understanding answer

    For slipping to occur F2<uR2 .
    What does F2 stand for . Friction or the force going to the left. Should it not be F2=uR2 as were looking for the least friction force possible so it's not greater than F2 .
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    Re: help understanding answer

    $F_1$ and $F_2$ are both forces of static friction equal in magnitude.

    $F_2 \color{red}{\le} \mu \cdot R_2$

    note that is less than or equal to ... equality occurring at the maximum force of static friction. $F_{2 \, max} = \mu \cdot R_2$.

    Have you conducted lab experiments to determine both coefficients of friction, or is that part of your study of mechanics lacking?
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    Re: help understanding answer

    Sorry I had it the other way around. So the friction force has to be less than or equal to uR2 otherwise it would be slipping. But the object is in equilibrium.
    I only do questions and I'm not at the level yet of doing this stuff experimentally.
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