Because I don't have enough rep to post images, the question is question 3 part iii) from here: http://www.mei.org.uk/files/papers/m209ju_jk32.pdf

The surface at the contact point B is rough. The reaction at B is: $R_{b}$ and the friction at B is: $F$ acting left, also $\tan(\alpha)=\frac{15}{8}$. I am trying to find the friction, $F$

Therefore if I resolve for the system vertically taking upwards as the positive direction I get: $$R_{b}-34\cos(\alpha)-85=0$$ Therefore $R_{b}=115$ therefore the reaction at B is $115N$

Taking clockwise moments about A gives: $$(34\times2.5)+(85\cos(\alpha)\times3)+(F\sin(\alpha)\times5)-(115\cos(\alpha)\times5)=0 $$ Where $\cos(\alpha)=\frac{15}{17}$ and $\sin(\alpha)=\frac{8}{17}$

Solving the equation for $F$, I get $F=\frac{671}{8} \approx 83.9$

But the answer says that $F = 7.4$, help on where I went wrong would be appreciated, thanks

  • $\begingroup$ Try this: Balance torque around B, to get the normal reaction at A. Using that, balance the net horizontal force to calculate the friction at B. Check to see what numbers that gives. $\endgroup$ – Siva May 27 '13 at 21:05
  • $\begingroup$ Wait, if $\tan(\alpha)>1$ then how can you have $\sin(\alpha)<\sqrt{2}$. You've flipped the values of sine and cosine. $\endgroup$ – Siva May 27 '13 at 21:14

The correct answer is $7.4$ only. Your $\cos()$ and $\sin()$ are interchanged.

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  • $\begingroup$ Yea, silly mistake, whoops $\endgroup$ – Johnmgee May 27 '13 at 22:20
  • $\begingroup$ :) Happens all the time. $\endgroup$ – Man May 27 '13 at 22:21

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