Answer to Question #140917 in Molecular Physics | Thermodynamics for Savannah

Question

A 22 kg dog takes advantage of its owner's absence to jump up on the furniture. It sits on a 1.2 meter long coffee table, 0.4 m from the right side. The table top has a mass of 5 kg and each leg is attached 0.1 m from the ends of the table. Find the force from the right and left legs of the coffee table. answer

Answer 1

Since the table is not moving, all forces are in equilibrium. The reaction forces in the legs of the table act upward, the force of gravity of the table top and dog are acting downward:

$$F_L+m_Tg+m_Dg+F_R=0.$$

Calculate the moments of forces created by the sitting dog around the right leg of the table. We have 5 kg spread along the table top (along 1.2 m). 5 kg is a uniform load, each small piece of the table has 5/1.2 of its mass.

$$-\frac{5}{1.2}\cdot0.1\cdot\frac{0.1}{2}+22\cdot0.3+\frac{5}{1.2}\cdot1.1\cdot\frac{1.1}{2}-F_L\cdot1=0,\\ F_L=9.1\text{ N}.$$

From the first equation, we have

$$F_R=255.5 \text{ N}.$$

Additionally, you may check that the sum of reaction forces in the legs is equal to the sum of loads, or

$$F_L+F_R=g(m_T+m_D).$$