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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 $\displaystyle \pi/2- \theta$ (since they are of the same length, angle $\displaystyle 2\theta$ is divided equally). Separate that force into horizontal and vertical components. In order that there be no slip, the friction force, $\displaystyle \mu$ 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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