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Hi, need help answering this...
A uniform rod AB of weight W is placed so as to learn against a step 12cm high, the foot of the rod A resting on horizontal ground. The angle of the rod to the horizontal is sininverse (4/5) and the rod is 20cm long. Find the coefficient of friction u at the ground and the step(given that they are equal) if the rod is in limiting equilibrium.
Ans u=o.362
I have drawn the reaction forces in the diagram
Have you considered calculating moments around the point on the step that the ladder leans against?Quote:
Originally Posted by Kai
Hmm, moments ? i dont understand how...Quote:
Originally Posted by mr fantastic
Maybe you know it as torque ....?Quote:
Originally Posted by Kai
I will describe this briefly.Quote:
Originally Posted by Kai
You have your extended FBD, but you haven't labeled two of your forces. There is a horizontal friction force at the base of the rod and a vertical one where the rod meets the step. Thus you have 4 unknown forces in the diagram.
Apply Newton's 2nd in each coordinate direction. That will give you two linearly independent equations. So you need two torque equations. Since nothing in the diagram is rotating we are free to choose an "axis of rotation" to be anywhere we like.
So I'd set an axis first where the rod meets the ground and calculate the sum of the torques about that axis. The sum of the torques is, of course, 0 Nm because there is no angular acceleration. Then put the axis at the point where the rod meets the step and do the sum of torques again.
This will give you four equations in four unknowns, which you can solve.
-Dan
Thanks for the help, i have a better idea ( actually didnt think at all about involving moments in it ), will try to do it asap