The unstretched length of spring AB is 3 m. If the block is held in the equilibrium position shown, determine the mass of the block at D.
Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
Solution:
Let us first determine the stretched length of spring AB. We can use Pythagorean theorem to figure this out.
Length of spring = 5 m
We know from the question that the unstretched length of the spring is 3 m. That would mean the spring stretched by 2 m (5 m – 3 m = 2m) after the block was hung at ring A.
We can now figure out the force of spring AB using Hook’s Law.
(Where F is force, k is the stiffness of the spring, and s is the stretch of the spring)
F\,=\,(30)(2)F\,=\,60 N
Let us now draw our free body diagram.
The angles were calculated using the inverse of tan (arctan).
\text{tan}^{-1}\left(\dfrac{3}{3}\right)\,=\,45^0
The orange angle was found by:
\text{tan}^{-1}\left(\dfrac{3}{4}\right)\,=\,36.87^0
The next step is to write our equations of equilibrium.
\rightarrow ^+\sum \text{F}_\text{x}\,=\,0
60\text{cos}\,(36.87^0)\,-\,F_{AC}\text{cos}\,(45^0)\,=\,0
(solve for F_{AC})
F_{AC}\,=\,67.88 N
+\uparrow \sum \text{F}_\text{y}\,=\,0
60\text{sin}\,(36.87^0)\,+\,F_{AC}\text{sin}\,(45^0)\,-\,W\,=\,0
(Substitute the value of F_{AC} we found and solve for W)
W\,=\,84 N
Show me the free body diagram
To figure out the mass, remember that W=mg.
(Where W is weight, m is mass, and g is the force of gravity)
m\,=\,\dfrac{84}{9.81}
m\,=\,8.56 kg
Final Answer:
mass = 8.56 kg
Hi, I have a question. Why don’t we just calculate the length of AB with square root of 3^2+3^2 ?
Thank you for answering!
So the lengths of the sides are 4m and 3 m. That’s what we have to use to calculate the length of AB. Maybe you misread the dimensions given? Let me know if I am misunderstanding your question.