Chapter 9, Problem 29P

To determine

The horizontal and vertical force components exerted by each hinge on the door.

Answer to Problem 29P

Solution:

The horizontal and vertical forces exerted on upper hinge are $\text{55}.\text{21 N and 63}.\text{7N}$.

The horizontal and vertical forces exerted on lower hinge are $\text{55}.\text{21 N and 63}.\text{7N}$.

Explanation of Solution

Given:

Mass of the door , $m=13\text{ kg}$

Formula used:

Equilibrium condition: the net force and torque would be zero.

  $\begin{array}{l}{\displaystyle \sum F=0}\\ {\displaystyle \sum \tau =0}\end{array}$

Calculation:

Let $A_x$ , $A_y$ are the horizontal and vertical forces at the upper hinge

Let $B_x$ , $B_y$ are the horizontal and vertical forces at the upper hinge

  $\begin{array}{c}{\displaystyle \sum F}=0\\ A_y+B_y-mg=0\\ A_y+A_y=mg\text{ (}{A}_y=B_y)\\ 2A_y=13(9.8)\\ A_y=63.7\text{N}\\ A_y=B_y=63.7\text{N}\end{array}$

Now taking sum of all the torques about hinge A

  $\begin{array}{c}{\displaystyle \sum \tau }=0\\ B_x\left(2.3-0.4-0.4\right)-mg\left(\frac{1.3}{2}\right)=0\\ 1.5B_x=13\left(9.8\right)\left(0.65\right)\\ B_x=55.21\text{N}\\ B_x=A_x=55.21\text{N}\end{array}$

Conclusion:

The horizontal and vertical forces exerted on upper hinge are $\text{55}.\text{21 N and 63}.\text{7N}$.

The horizontal and vertical forces exerted on lower hinge are $\text{55}.\text{21 N and 63}.\text{7N}$.