The roof has the shape of equlateral triange?

Question

Horizontal thin rod of mass 10kg is hanging under the roof supported by friction μ=0.600 from the sides of the roof, and nothing else:

- - - -/^\
- - -/- - -\
- -/=rod=\
-/- - - - - - -\

What is tension T in the rod?

Answer 1

I assume you want the minimum tension (?) or compression force F to support the rod.
The friction force Ff = μ (F cos30 - 0.5 m g sin30).
The tangential force opposing Ff (due to m/2 and F) Ft = F sin30 + 0.5 m g cos30.
Solving, F = 2916 N.
As answer 2 states, it might not be possible to support the bar because the geometry prevents the friction force from exceeding the downward force due to the bar force. Indeed we are close to this point with the angles given. For the given value of μ, there is a critical angle from the vertical above which no amount of force will hold the load, since μ cos(theta) < sin(theta) thus theta > arctan(mu). This angle is about 30.96 deg.

Answer 2

Assume the roof to form an equilateral triangle and that the ends of the rod are cut to meet the roof such that the surfaces are parallel. The rod is fully supported by the friction from the two ends and each end must support half of the weight. Keep in mind that the component of frictional force that is anti-parallel to gravity is only sin(60) of the total force provided by the friction and that the normal force between the rod and the roof is only a cos(30) of the tension force. You must then show that a sufficiently large tension is capable of causing a large enough normal force and therefore a large enough frictional force to overcome gravity. It may be that there is no such force, for a component of the normal force is also acting with gravity, in which case the rod would fall and the tension is zero. If there is a non zero solution, there are likely to be many, in other words a range of possible tensions. The above information should be enough to help you set up your analysis, I am sure you can do the actual number crunching.

Answer 3

10kg * 9.81m/s^2 = 98.1 N
98.1N devided equally on each side of the rod is 49.05N
49.05N / 0.600 = 81.75N normal friction
So the tension in the rod is of 81.75N due to the action reaction law