Question

Q. Three like horizontal forces of 10N, 20N, and 10N act on a vertical rod at A, B, C. If AB = BC = 20 mm. The resultant force couple system at A is A. 40 N, 800 Nmm B. 0 N, 400 Nmm C. 20 N, 200 Nmm D. None of these Ans. A​

Answer 1

The answer will be option A.

Explanation:

A) 40 N, 800 Nmm.

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Answer 2

Answer:

Option D is correct answer

Explanation:

From the geometry of the figure, we find that the triangle $A B C$ is a right-angled triangle, in which side $AB=BC=20 \mathrm{~mm} .$

$AC=\sqrt{AB^{2} +BC^{2} } \\=\sqrt{20^{2}+20^{2} } \\=\sqrt{800}\\ =28.28$

Therefore

$&{Sin} \theta\\=\frac{20}{28}\\=0.71$

$&\cos \theta\\=\frac{20}{28}\\=0.71$

Resolving all the forces horizontally (i.e., along $A B$ ),

$\begin{aligned}\sum H=10-30 \cos \theta \\=10-(30 \times 0.71)\\=-11.3 \mathrm{~N}\end{aligned}$

and now resolving all the forces vertically (i.e., along $B C$ )

$\begin{aligned}\sum V=20-30 \sin \theta \\=20-(30 \times 0.71)\\=1.3 \mathrm{~N}\end{aligned}$

We know that magnitude of the resultant force,

$\mathrm{R}=\sqrt{\left(\sum H\right)^{2}+\left(\sum V\right)^{2}}\\=\sqrt{-11.3^{2}+1.3^{2}}\\=17.22 \mathrm{~N}$