The resultant of these forces OP, OQ, OR, OS and OT is approximately ...... N. [Take, {3}=1.7, {2}=1.4 and given {i} and {j} unit vectors along X, Y axis [27 Aug 2021 Shift 1]

Correct Answer is : 9.25i^+5j^9.25 {i}+5 {j}9.25i^+5j^

Correct Answer is : 9.25i^+5j^9.25 {i}+5 {j}9.25i^+5j^

Test Index

Vector Algebra

Section: Physics

Question : 30 of 47

Marks: +1, -0

\sqrt{3}=1.7, \sqrt{2}=1.4

\hat{i}

\hat{j}

Solution:

From given figure, we can write all the forces in vector notation as

OP=(20 \sin 30^{\circ} \hat{i}+20 \cos 30^{\circ} \hat{j}) \mathrm{N}

=(10 \hat{i}+17.32 \hat{j}) \mathrm{N}

OQ=(10 \cos 30^{\circ} \hat{i}+10 \sin 30^{\circ} \hat{j}) \mathrm{N}

=(8.66 \hat{i}+5 \hat{j}) \mathrm{N}

OR=(20 \cos 45^{\circ} \hat{i}-20 \sin 45^{\circ} \hat{j}) \mathrm{N}

=(14.14 \hat{i}-14.14 \hat{j}) \mathrm{N}

OS=(-15 \cos 45^{\circ} \hat{i}-15 \sin 45^{\circ} \hat{j}) \mathrm{N}

=(-10.60 \hat{i}-10.60 \hat{j}) \mathrm{N}

OT=(-15 \sin 60^{\circ} \hat{i}+15 \cos 60^{\circ} \hat{j}) \mathrm{N}

=(-12.99 \hat{i}+7.5 \hat{j}) \mathrm{N}

The resultant of these forces is given as

F = OP + OQ + OR + OS + OT

F=[(10 \hat{i}+17.32 \hat{j})+(8.66 \hat{i}+5 \hat{j})+(14.14 \hat{i}-14.14 \hat{i})+(-10.60 \hat{i}-10.60 \hat{j})+(-12.99 \hat{i}+7.5 \hat{j})]

F=9.21 \hat{i}+5.08 \hat{j} \mathrm{N}

F \approx 9.25 \hat{i}+5 \hat{j} \mathrm{N}

Thus, the resultant force is

F\approx 9.25 \hat{i}+5 \hat{j} \mathrm{N}

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