0

"A uniform beam, AB, of mass 14kg and length 15m, rests with A on rough, horizontal ground. The beam is supported by a smooth peg at C, where AC=13m, so that it makes an angle of 18 degrees with the horizontal. The beam is on the point of slipping.

Find the coefficient of friction between the ground and the beam at A."

I calculated the reaction forces at A and C. I then equated the frictional force, F, at A to be the sum of the horizontal components of both the weight of AB, mg, and the reaction force at C (before dividing F by the reaction force at A), b however the answer given in the back of the textboot equates F to the latter alone.

Am I correct to consider a horizontal component of mg? I calculated it to be 14gsin(18)cos(18), by first calculating the component parallel to the beam using a right angle triangle, and then drawing a second right angle triangle with the parallel component of the weight as the hypotenuse, and calculating the horizontal component of that.

Photograph of my working

I shall be grateful for any assistance.

Edit 1: After initially saying I had been unable to, I've managed to add a picture (on Firefox for Android, I was not informed as to why my image could not be added, but once I tried of Firefox for Windows, I discovered the image was too large).

Miles Bardan
  • 1
  • A picture should be very useful, I hope to have correctly interpreted the OP in my answer, check it and tell me if it can help – user Feb 11 '18 at 23:31

1 Answers1

1

HINT

Indicating with

  • W=mg weight of the beam
  • AB=15m length of the beam
  • AC=13m distance AC
  • $V_C$ reaction in C (perpendicular to the beam)
  • $V_A$ vertical reaction in A (up)
  • $H_A$ horizontal reaction in A

By moment equation with respect to A we obtain

$$W\cdot \frac12 AB\cdot \cos 18°=V_C\cdot AC \implies V_C=W\cdot \frac12 \frac{AB}{AC}\cdot \cos 18°$$

then by vertical equilibrium in $A$ we obtain

$$V_A+V_C \cos 18°=W\implies V_A=W-V_C\cos 18°$$

and by horizontal equilibrium in $A$ we obtain

$$H_A=V_C \sin 18°$$

finally the coefficient of friction is given by

$$H_A=\mu\cdot V_A\implies \mu=\frac{H_A}{V_A}$$

enter image description here

user
  • 154,566
  • Thank you gimusi :) The answer you've given is the same as the one given in the textbook, and the working you've done is the same as what I did except that I also considered a horizontal component of W, and this is this part I don't understand.

    I've now uploaded a photograph of the part of my working which is relevant to my question. I'd be very grateful if you'd take a look at it, and tell me whether my calculation of a horizontal component for the weight makes any sense or not, and if not why not.

    Thank you again!

     – Miles Bardan Feb 12 '18 at 20:10
  • @MilesBardan I've added a sketch with the force to be considered for equilibrium, as you can note W has only vertical component, thus it doesn't give any contribution for equilibrium in horizontal direction. – user Feb 12 '18 at 20:54