2003 Paper 2 Q9

Year: 2003
Paper: 2
Question Number: 9

Course: LFM Pure and Mechanics
Section: Moments

Difficulty: 1600.0 Banger: 1484.0

Problem

\(AB\) is a uniform rod of weight \(W\,\). The point \(C\) on \(AB\) is such that \(AC>CB\,\). The rod is in contact with a rough horizontal floor at \(A\,\) and with a cylinder at \(C\,\). The cylinder is fixed to the floor with its axis horizontal. The rod makes an angle \({\alpha}\) with the horizontal and lies in a vertical plane perpendicular to the axis of the cylinder. The coefficient of friction between the rod and the floor is \(\tan \lambda_1\) and the coefficient of friction between the rod and the cylinder is \(\tan \lambda_2\,\). Show that if friction is limiting both at \(A\) and at \(C\), and \({\alpha} \ne {\lambda}_2 - {\lambda}_1\,\), then the frictional force acting on the rod at \(A\) has magnitude $$ \frac{ W\sin {\lambda}_1 \, \sin({\alpha}-{\lambda}_2)} {\sin ({\alpha}+{\lambda}_1-{\lambda}_2)} \;.$$ %and that %$$ %p=\frac{\cos{\alpha} \, \sin({\alpha}+{\lambda}_1-{\lambda}_2)} %{2\cos{\lambda}_1 \, \sin {\lambda}_2}\;. %$$

No solution available for this problem.

Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1484.0

Banger Comparisons: 1

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Problem source
$AB$ is a uniform rod of weight $W\,$.
The point $C$  on $AB$ is such that $AC>CB\,$. The 
rod is in contact with a rough horizontal floor  at $A\,$ 
and with a cylinder at $C\,$. The cylinder   
is fixed to the floor with its axis horizontal.
The rod makes an angle ${\alpha}$ with 
the horizontal and lies in a vertical plane perpendicular to
the axis of the cylinder.
The coefficient  of friction between the rod
and the floor  is $\tan \lambda_1$ and the coefficient of friction
between the rod and the 
cylinder is $\tan \lambda_2\,$.  
Show that if friction is limiting 
both at $A$ and at $C$, and ${\alpha} \ne {\lambda}_2 - {\lambda}_1\,$, 
then the frictional force acting on the rod 
at $A$ has magnitude 
$$
\frac{ W\sin {\lambda}_1  \, \sin({\alpha}-{\lambda}_2)}
{\sin ({\alpha}+{\lambda}_1-{\lambda}_2)}
\;.$$
%and that
%$$
%p=\frac{\cos{\alpha} \, \sin({\alpha}+{\lambda}_1-{\lambda}_2)}
%{2\cos{\lambda}_1 \, \sin {\lambda}_2}\;.
%$$