The problem is as follows:
The figure from below shows a homogeneous triangular shaped plate. Find the tension of the wire which is holding the plate to the ceiling. The mass of the plate is $15\,kg$. The lenght indicated with two paralell lines indicates that the mentioned sides are the same. You may consider that the acceleration due gravity is $10\frac{m}{s^2}$.
The alternatives are as follows:
$\begin{array}{ll} 1.&50\,N\\ 2.&70\,N\\ 3.&90\,N\\ 4.&100\,N\\ 5.&120\,N\\ \end{array}$
I'm confused about this question as no other information is given. How exactly am I supposed to find the tension in the wire as the geometry of the object doesn't really provide much information other than being a triangle. It doesn't say if it is equilateral or anything and there isn't any clue of what sort it can be. I'm assuming that the way to approach this problem would be using the condition of equilibrium as as the sum of torques would be equated to zero. But how to establish the distance to the center of this figure?, can somebody help me with this?.
You're right that the solution lies in the torques.
We need to assume that there's no horizontal static frictional force at the point of contact on the floor. Then the only torques are from the gravitational force and from the upward force exerted by the floor. The gravitational force can be supposed to apply to the centre of mass, and the horizontal distance of the centre of mass from the vertical line of suspension is $\frac13$ that of the supported tip. Thus the gravitational force must be $3$ times the upward force at the tip, so $\frac23$ of the gravitational force, or $100N$, are carried as tension by the rope.