Answer:
68.829 N
Explanation:
The given parameters are:
Weight of ladder, [tex]W_l[/tex] = 120 N
Weight of object, [tex]W_o[/tex] = 98 N
Angle, [tex]\theta[/tex] = 53°
And we also know that, while
Length of ladder = L
Distance the object is placed = L/3
If we apply translational equilibrium horizontally, then [tex]T-N=0[/tex], so [tex]T = N[/tex]
If we apply rotational equilibrium about the distance the object is placed, then
[tex]W_o\frac{L}{3}cos(\theta) + W_l\frac{L}{2}cos(\theta) - NLsin(\theta) = 0\\2NLsin(\theta) = 2W_o\frac{L}{3}cos(\theta)+W_lLcos(\theta)\\2Tsin(\theta) = \frac{2}{3}W_ocos(\theta)+W_lLcos(\theta)\\T = \frac{1}{3}W_ocot(\theta) + \frac{1}{2}W_lcot(\theta)\\T = \frac{1}{3} * 98 * cot(53) + \frac{1}{2} * 120 * cot(53) = 69.829[/tex]
