A uniform girder AB of length $10m$ and weight $20000N$ rests with its end $A$ on the rough horizontal floor of a factory and with its other end $B$ supported by a cable. The girder makes an angle $α$ with the floor, and the cable is inclined at an angle $β$ to the horizontal. The girder is in limiting equilibrium, and the coefficient of friction between the girder and the floor is $μ$.

We have $tanβ$ = $\frac{20000 - R}{μR}$

where $R$ is the magnitude of the normal force acting on the girder at $A$.

Given $tanα = \frac{3}{4}$ and $tanβ = 5$

Show that:

a) $μ = \frac{2}{7}$

b) find the tension in the cable. (should be $12000N$)


Resolving horizontally:

$F = Tcosβ$ , where $F$ is the frictional force at $A$ and $T$ is the tension on the cable. Also $F = μR$ (limiting equilibrium).

Resolving vertically:

$R + Tsinβ = W$

What is the relationship of $tanα = \frac{3}{4}$ and $tanβ = 5$ to $μ$?


HINT...you can take moments at, for example, the point of contact of the girder and the floor and obtain $$20000\times5\cos \alpha=T\times10\sin(\beta-\alpha)$$

I hope this helps.

  • $\begingroup$ It does help, thanks. $sin(β-α)=sinβsinα-cosβsinα$ = $\frac{5}{sqrt26} * \frac{4}{5} - \frac{1}{sqrt26}*\frac{3}{5} = \frac{17}{5*sqrt26}$ $\endgroup$
    – J132
    Sep 8 '15 at 18:34
  • $\begingroup$ You're welcome. $\endgroup$ Sep 8 '15 at 18:40
  • $\begingroup$ It does help, thanks. $sin(β-α)=sinβsinα-cosβsinα$ = $\frac{5}{sqrt26} * \frac{4}{5} - \frac{1}{sqrt26}*\frac{3}{5} = \frac{17}{5*sqrt26}$ so $T = \frac{40000*sqrt26}{17} = 12000N $ to the nearest whole. Now $μ= \frac{F}{R} = \frac{Tcosβ}{W-Tsinβ} = \frac{\frac{40000*sqrt26}{17}*\frac{1}{sqrt26}}{20000 - \frac{40000}{17} * \frac{5}{sqrt26}}$ = ${\frac{40000}{17}*\frac{17}{140000}}$ = $\frac{2}{7}$ $\endgroup$
    – J132
    Sep 8 '15 at 18:54
  • $\begingroup$ Should say '$12000N$ to the nearest thousand'. $\endgroup$
    – J132
    Sep 8 '15 at 21:08

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