145988
One of the rectangular components of a force \(40 \mathrm{~N}\) is \(20 \sqrt{3} \mathrm{~N}\). What is the other rectangular component?
1 \(10 \mathrm{~N}\)
2 \(20 \mathrm{~N}\)
3 \(30 \mathrm{~N}\)
4 \(25 \mathrm{~N}\)
Explanation:
B Given, Rectangular component of \(40 \mathrm{~N}\) force \(=20 \sqrt{3} \mathrm{~N}\) Let, \(\quad F_{x}=20 \sqrt{3}, \quad F=40 \mathrm{~N}\) \(\mathrm{F}_{\mathrm{y}}=?\) We know that, \(\mathrm{F}^{2}=\mathrm{F}_{\mathrm{x}}^{2}+\mathrm{F}_{\mathrm{y}}^{2}\) \((40)^{2}=(20 \sqrt{3})^{2}+\mathrm{F}_{\mathrm{y}}^{2}\) \(1600=1200+\mathrm{F}_{\mathrm{y}}^{2}\) \(\mathrm{~F}_{\mathrm{y}}^{2}=400=20 \times 20\) \(\mathrm{~F}_{\mathrm{y}}=20\) Hence, other component is \(=20 \mathrm{~N}\)
AP EAMCET-19.08.2021
Laws of Motion
145989
When a lift of mass \(800 \mathrm{~kg}\) is ascending with an acceleration of \(5 \mathrm{~m}^{-2}\), the tension in its cable will be - \(\left(\right.\) take \(g=10 \mathrm{~m} . \mathrm{s}^{-2}\) )
1 \(6000 \mathrm{~N}\)
2 \(12000 \mathrm{~N}\)
3 \(4000 \mathrm{~N}\)
4 \(50 \mathrm{~N}\)
Explanation:
B Given that, \(\mathrm{m}=800 \mathrm{~kg} . \quad \mathrm{a}=5 \mathrm{~m} / \mathrm{sec}^{2} \quad \mathrm{~g}=10 \mathrm{~m} / \mathrm{sec}^{2}\) Let the tension in the cable be \(T\), From eq. of motion for lift \(\mathrm{T}-\mathrm{mg}=\mathrm{ma}\) \(\mathrm{T}=\mathrm{ma}+\mathrm{mg}\) \(\mathrm{T}=\mathrm{m}(\mathrm{g}+\mathrm{a})\) \(\mathrm{T}=800(10+5)\) \(\mathrm{T}=12000 \mathrm{~N}\) Tension in its cable will be \(12000 \mathrm{~N}\).
AP EAMCET-06.09.2021
Laws of Motion
145990
Three forces of magnitude \(6 \mathrm{~N}, 6 \mathrm{~N}\) and \(\sqrt{72} \mathrm{~N}\) act at a corner of a cube along three edges of a cube, as shown in the figure. The resultant of the three forces is
1 \(12 \mathrm{~N}\) along \(\mathrm{OM}\)
2 \(18 \mathrm{~N}\) along \(\mathrm{OA}\)
3 \(18 \mathrm{~N}\) along \(\mathrm{OC}\)
4 \(12 \mathrm{~N}\) along \(\mathrm{OE}\)
Explanation:
D Given, Force along \((\mathrm{OA})=6 \mathrm{~N}\) Force along \((\mathrm{OC})=6 \mathrm{~N}\) Force along \((\mathrm{OG})=\sqrt{72} \mathrm{~N}\) The Resultant of these forces along \(\mathrm{OE}\) is, \(\mathrm{R}=\sqrt{(\mathrm{OA})^{2}+(\mathrm{OC})^{2}+(\mathrm{OG})^{2}}\) \(\mathrm{R}=\sqrt{6^{2}+6^{2}+(\sqrt{72})^{2}}\) \(\mathrm{R}=\sqrt{36+36+72}\) \(\mathrm{R}=\sqrt{144}\) \(\mathrm{R}=12 \mathrm{~N} \text { along } \mathrm{OE}\) Hence, the resultant of the three forces is \(12 \mathrm{~N}\) along OE.
AP EAMCET-06.09.2021
Laws of Motion
145991
An object is in equilibrium when four concurrent forces, acting in the same plane, are in the directions shown in the figure. Find the magnitudes of \(F_{1}\) and \(F_{2}\).
B Given, object is in equilibrium. (i) For all forces acting along the \(\mathrm{x}\)-axis \(\left(\sum \mathrm{F}_{\mathrm{x}}=0\right)\) \(\therefore 8+4 \cos 60^{\circ}-\mathrm{F}_{2} \cos 30^{\circ}=0\) \(8+2-\mathrm{F}_{2} \frac{\sqrt{3}}{2}=0\) \(\mathrm{~F}_{2}=\frac{20}{\sqrt{3}} \mathrm{~N}\) Now, (ii) For the force acting in along the y-axis \(\left(\sum \mathrm{F}_{\mathrm{y}}=0\right)\) \(\therefore \mathrm{F}_{1}+4 \sin 60^{\circ}-\mathrm{F}_{2} \sin 30^{\circ}=0\) \(\mathrm{~F}_{1}+\frac{4 \sqrt{3}}{2}-\frac{\mathrm{F}_{2}}{2}=0\) \(\mathrm{~F}_{1}=\frac{\mathrm{F}_{2}}{2}-2 \sqrt{3}=\frac{10}{\sqrt{3}}-2 \sqrt{3}\) \(\mathrm{~F}_{1}=\frac{4}{\sqrt{3}} \mathrm{~N}\) Hence, magnitude of \(\mathrm{F}_{1}=\frac{4}{\sqrt{3}} \mathrm{~N}\) and \(\text { Magnitude of } F_{2}=\frac{20}{\sqrt{3}} \mathrm{~N}\)
145988
One of the rectangular components of a force \(40 \mathrm{~N}\) is \(20 \sqrt{3} \mathrm{~N}\). What is the other rectangular component?
1 \(10 \mathrm{~N}\)
2 \(20 \mathrm{~N}\)
3 \(30 \mathrm{~N}\)
4 \(25 \mathrm{~N}\)
Explanation:
B Given, Rectangular component of \(40 \mathrm{~N}\) force \(=20 \sqrt{3} \mathrm{~N}\) Let, \(\quad F_{x}=20 \sqrt{3}, \quad F=40 \mathrm{~N}\) \(\mathrm{F}_{\mathrm{y}}=?\) We know that, \(\mathrm{F}^{2}=\mathrm{F}_{\mathrm{x}}^{2}+\mathrm{F}_{\mathrm{y}}^{2}\) \((40)^{2}=(20 \sqrt{3})^{2}+\mathrm{F}_{\mathrm{y}}^{2}\) \(1600=1200+\mathrm{F}_{\mathrm{y}}^{2}\) \(\mathrm{~F}_{\mathrm{y}}^{2}=400=20 \times 20\) \(\mathrm{~F}_{\mathrm{y}}=20\) Hence, other component is \(=20 \mathrm{~N}\)
AP EAMCET-19.08.2021
Laws of Motion
145989
When a lift of mass \(800 \mathrm{~kg}\) is ascending with an acceleration of \(5 \mathrm{~m}^{-2}\), the tension in its cable will be - \(\left(\right.\) take \(g=10 \mathrm{~m} . \mathrm{s}^{-2}\) )
1 \(6000 \mathrm{~N}\)
2 \(12000 \mathrm{~N}\)
3 \(4000 \mathrm{~N}\)
4 \(50 \mathrm{~N}\)
Explanation:
B Given that, \(\mathrm{m}=800 \mathrm{~kg} . \quad \mathrm{a}=5 \mathrm{~m} / \mathrm{sec}^{2} \quad \mathrm{~g}=10 \mathrm{~m} / \mathrm{sec}^{2}\) Let the tension in the cable be \(T\), From eq. of motion for lift \(\mathrm{T}-\mathrm{mg}=\mathrm{ma}\) \(\mathrm{T}=\mathrm{ma}+\mathrm{mg}\) \(\mathrm{T}=\mathrm{m}(\mathrm{g}+\mathrm{a})\) \(\mathrm{T}=800(10+5)\) \(\mathrm{T}=12000 \mathrm{~N}\) Tension in its cable will be \(12000 \mathrm{~N}\).
AP EAMCET-06.09.2021
Laws of Motion
145990
Three forces of magnitude \(6 \mathrm{~N}, 6 \mathrm{~N}\) and \(\sqrt{72} \mathrm{~N}\) act at a corner of a cube along three edges of a cube, as shown in the figure. The resultant of the three forces is
1 \(12 \mathrm{~N}\) along \(\mathrm{OM}\)
2 \(18 \mathrm{~N}\) along \(\mathrm{OA}\)
3 \(18 \mathrm{~N}\) along \(\mathrm{OC}\)
4 \(12 \mathrm{~N}\) along \(\mathrm{OE}\)
Explanation:
D Given, Force along \((\mathrm{OA})=6 \mathrm{~N}\) Force along \((\mathrm{OC})=6 \mathrm{~N}\) Force along \((\mathrm{OG})=\sqrt{72} \mathrm{~N}\) The Resultant of these forces along \(\mathrm{OE}\) is, \(\mathrm{R}=\sqrt{(\mathrm{OA})^{2}+(\mathrm{OC})^{2}+(\mathrm{OG})^{2}}\) \(\mathrm{R}=\sqrt{6^{2}+6^{2}+(\sqrt{72})^{2}}\) \(\mathrm{R}=\sqrt{36+36+72}\) \(\mathrm{R}=\sqrt{144}\) \(\mathrm{R}=12 \mathrm{~N} \text { along } \mathrm{OE}\) Hence, the resultant of the three forces is \(12 \mathrm{~N}\) along OE.
AP EAMCET-06.09.2021
Laws of Motion
145991
An object is in equilibrium when four concurrent forces, acting in the same plane, are in the directions shown in the figure. Find the magnitudes of \(F_{1}\) and \(F_{2}\).
B Given, object is in equilibrium. (i) For all forces acting along the \(\mathrm{x}\)-axis \(\left(\sum \mathrm{F}_{\mathrm{x}}=0\right)\) \(\therefore 8+4 \cos 60^{\circ}-\mathrm{F}_{2} \cos 30^{\circ}=0\) \(8+2-\mathrm{F}_{2} \frac{\sqrt{3}}{2}=0\) \(\mathrm{~F}_{2}=\frac{20}{\sqrt{3}} \mathrm{~N}\) Now, (ii) For the force acting in along the y-axis \(\left(\sum \mathrm{F}_{\mathrm{y}}=0\right)\) \(\therefore \mathrm{F}_{1}+4 \sin 60^{\circ}-\mathrm{F}_{2} \sin 30^{\circ}=0\) \(\mathrm{~F}_{1}+\frac{4 \sqrt{3}}{2}-\frac{\mathrm{F}_{2}}{2}=0\) \(\mathrm{~F}_{1}=\frac{\mathrm{F}_{2}}{2}-2 \sqrt{3}=\frac{10}{\sqrt{3}}-2 \sqrt{3}\) \(\mathrm{~F}_{1}=\frac{4}{\sqrt{3}} \mathrm{~N}\) Hence, magnitude of \(\mathrm{F}_{1}=\frac{4}{\sqrt{3}} \mathrm{~N}\) and \(\text { Magnitude of } F_{2}=\frac{20}{\sqrt{3}} \mathrm{~N}\)
145988
One of the rectangular components of a force \(40 \mathrm{~N}\) is \(20 \sqrt{3} \mathrm{~N}\). What is the other rectangular component?
1 \(10 \mathrm{~N}\)
2 \(20 \mathrm{~N}\)
3 \(30 \mathrm{~N}\)
4 \(25 \mathrm{~N}\)
Explanation:
B Given, Rectangular component of \(40 \mathrm{~N}\) force \(=20 \sqrt{3} \mathrm{~N}\) Let, \(\quad F_{x}=20 \sqrt{3}, \quad F=40 \mathrm{~N}\) \(\mathrm{F}_{\mathrm{y}}=?\) We know that, \(\mathrm{F}^{2}=\mathrm{F}_{\mathrm{x}}^{2}+\mathrm{F}_{\mathrm{y}}^{2}\) \((40)^{2}=(20 \sqrt{3})^{2}+\mathrm{F}_{\mathrm{y}}^{2}\) \(1600=1200+\mathrm{F}_{\mathrm{y}}^{2}\) \(\mathrm{~F}_{\mathrm{y}}^{2}=400=20 \times 20\) \(\mathrm{~F}_{\mathrm{y}}=20\) Hence, other component is \(=20 \mathrm{~N}\)
AP EAMCET-19.08.2021
Laws of Motion
145989
When a lift of mass \(800 \mathrm{~kg}\) is ascending with an acceleration of \(5 \mathrm{~m}^{-2}\), the tension in its cable will be - \(\left(\right.\) take \(g=10 \mathrm{~m} . \mathrm{s}^{-2}\) )
1 \(6000 \mathrm{~N}\)
2 \(12000 \mathrm{~N}\)
3 \(4000 \mathrm{~N}\)
4 \(50 \mathrm{~N}\)
Explanation:
B Given that, \(\mathrm{m}=800 \mathrm{~kg} . \quad \mathrm{a}=5 \mathrm{~m} / \mathrm{sec}^{2} \quad \mathrm{~g}=10 \mathrm{~m} / \mathrm{sec}^{2}\) Let the tension in the cable be \(T\), From eq. of motion for lift \(\mathrm{T}-\mathrm{mg}=\mathrm{ma}\) \(\mathrm{T}=\mathrm{ma}+\mathrm{mg}\) \(\mathrm{T}=\mathrm{m}(\mathrm{g}+\mathrm{a})\) \(\mathrm{T}=800(10+5)\) \(\mathrm{T}=12000 \mathrm{~N}\) Tension in its cable will be \(12000 \mathrm{~N}\).
AP EAMCET-06.09.2021
Laws of Motion
145990
Three forces of magnitude \(6 \mathrm{~N}, 6 \mathrm{~N}\) and \(\sqrt{72} \mathrm{~N}\) act at a corner of a cube along three edges of a cube, as shown in the figure. The resultant of the three forces is
1 \(12 \mathrm{~N}\) along \(\mathrm{OM}\)
2 \(18 \mathrm{~N}\) along \(\mathrm{OA}\)
3 \(18 \mathrm{~N}\) along \(\mathrm{OC}\)
4 \(12 \mathrm{~N}\) along \(\mathrm{OE}\)
Explanation:
D Given, Force along \((\mathrm{OA})=6 \mathrm{~N}\) Force along \((\mathrm{OC})=6 \mathrm{~N}\) Force along \((\mathrm{OG})=\sqrt{72} \mathrm{~N}\) The Resultant of these forces along \(\mathrm{OE}\) is, \(\mathrm{R}=\sqrt{(\mathrm{OA})^{2}+(\mathrm{OC})^{2}+(\mathrm{OG})^{2}}\) \(\mathrm{R}=\sqrt{6^{2}+6^{2}+(\sqrt{72})^{2}}\) \(\mathrm{R}=\sqrt{36+36+72}\) \(\mathrm{R}=\sqrt{144}\) \(\mathrm{R}=12 \mathrm{~N} \text { along } \mathrm{OE}\) Hence, the resultant of the three forces is \(12 \mathrm{~N}\) along OE.
AP EAMCET-06.09.2021
Laws of Motion
145991
An object is in equilibrium when four concurrent forces, acting in the same plane, are in the directions shown in the figure. Find the magnitudes of \(F_{1}\) and \(F_{2}\).
B Given, object is in equilibrium. (i) For all forces acting along the \(\mathrm{x}\)-axis \(\left(\sum \mathrm{F}_{\mathrm{x}}=0\right)\) \(\therefore 8+4 \cos 60^{\circ}-\mathrm{F}_{2} \cos 30^{\circ}=0\) \(8+2-\mathrm{F}_{2} \frac{\sqrt{3}}{2}=0\) \(\mathrm{~F}_{2}=\frac{20}{\sqrt{3}} \mathrm{~N}\) Now, (ii) For the force acting in along the y-axis \(\left(\sum \mathrm{F}_{\mathrm{y}}=0\right)\) \(\therefore \mathrm{F}_{1}+4 \sin 60^{\circ}-\mathrm{F}_{2} \sin 30^{\circ}=0\) \(\mathrm{~F}_{1}+\frac{4 \sqrt{3}}{2}-\frac{\mathrm{F}_{2}}{2}=0\) \(\mathrm{~F}_{1}=\frac{\mathrm{F}_{2}}{2}-2 \sqrt{3}=\frac{10}{\sqrt{3}}-2 \sqrt{3}\) \(\mathrm{~F}_{1}=\frac{4}{\sqrt{3}} \mathrm{~N}\) Hence, magnitude of \(\mathrm{F}_{1}=\frac{4}{\sqrt{3}} \mathrm{~N}\) and \(\text { Magnitude of } F_{2}=\frac{20}{\sqrt{3}} \mathrm{~N}\)
145988
One of the rectangular components of a force \(40 \mathrm{~N}\) is \(20 \sqrt{3} \mathrm{~N}\). What is the other rectangular component?
1 \(10 \mathrm{~N}\)
2 \(20 \mathrm{~N}\)
3 \(30 \mathrm{~N}\)
4 \(25 \mathrm{~N}\)
Explanation:
B Given, Rectangular component of \(40 \mathrm{~N}\) force \(=20 \sqrt{3} \mathrm{~N}\) Let, \(\quad F_{x}=20 \sqrt{3}, \quad F=40 \mathrm{~N}\) \(\mathrm{F}_{\mathrm{y}}=?\) We know that, \(\mathrm{F}^{2}=\mathrm{F}_{\mathrm{x}}^{2}+\mathrm{F}_{\mathrm{y}}^{2}\) \((40)^{2}=(20 \sqrt{3})^{2}+\mathrm{F}_{\mathrm{y}}^{2}\) \(1600=1200+\mathrm{F}_{\mathrm{y}}^{2}\) \(\mathrm{~F}_{\mathrm{y}}^{2}=400=20 \times 20\) \(\mathrm{~F}_{\mathrm{y}}=20\) Hence, other component is \(=20 \mathrm{~N}\)
AP EAMCET-19.08.2021
Laws of Motion
145989
When a lift of mass \(800 \mathrm{~kg}\) is ascending with an acceleration of \(5 \mathrm{~m}^{-2}\), the tension in its cable will be - \(\left(\right.\) take \(g=10 \mathrm{~m} . \mathrm{s}^{-2}\) )
1 \(6000 \mathrm{~N}\)
2 \(12000 \mathrm{~N}\)
3 \(4000 \mathrm{~N}\)
4 \(50 \mathrm{~N}\)
Explanation:
B Given that, \(\mathrm{m}=800 \mathrm{~kg} . \quad \mathrm{a}=5 \mathrm{~m} / \mathrm{sec}^{2} \quad \mathrm{~g}=10 \mathrm{~m} / \mathrm{sec}^{2}\) Let the tension in the cable be \(T\), From eq. of motion for lift \(\mathrm{T}-\mathrm{mg}=\mathrm{ma}\) \(\mathrm{T}=\mathrm{ma}+\mathrm{mg}\) \(\mathrm{T}=\mathrm{m}(\mathrm{g}+\mathrm{a})\) \(\mathrm{T}=800(10+5)\) \(\mathrm{T}=12000 \mathrm{~N}\) Tension in its cable will be \(12000 \mathrm{~N}\).
AP EAMCET-06.09.2021
Laws of Motion
145990
Three forces of magnitude \(6 \mathrm{~N}, 6 \mathrm{~N}\) and \(\sqrt{72} \mathrm{~N}\) act at a corner of a cube along three edges of a cube, as shown in the figure. The resultant of the three forces is
1 \(12 \mathrm{~N}\) along \(\mathrm{OM}\)
2 \(18 \mathrm{~N}\) along \(\mathrm{OA}\)
3 \(18 \mathrm{~N}\) along \(\mathrm{OC}\)
4 \(12 \mathrm{~N}\) along \(\mathrm{OE}\)
Explanation:
D Given, Force along \((\mathrm{OA})=6 \mathrm{~N}\) Force along \((\mathrm{OC})=6 \mathrm{~N}\) Force along \((\mathrm{OG})=\sqrt{72} \mathrm{~N}\) The Resultant of these forces along \(\mathrm{OE}\) is, \(\mathrm{R}=\sqrt{(\mathrm{OA})^{2}+(\mathrm{OC})^{2}+(\mathrm{OG})^{2}}\) \(\mathrm{R}=\sqrt{6^{2}+6^{2}+(\sqrt{72})^{2}}\) \(\mathrm{R}=\sqrt{36+36+72}\) \(\mathrm{R}=\sqrt{144}\) \(\mathrm{R}=12 \mathrm{~N} \text { along } \mathrm{OE}\) Hence, the resultant of the three forces is \(12 \mathrm{~N}\) along OE.
AP EAMCET-06.09.2021
Laws of Motion
145991
An object is in equilibrium when four concurrent forces, acting in the same plane, are in the directions shown in the figure. Find the magnitudes of \(F_{1}\) and \(F_{2}\).
B Given, object is in equilibrium. (i) For all forces acting along the \(\mathrm{x}\)-axis \(\left(\sum \mathrm{F}_{\mathrm{x}}=0\right)\) \(\therefore 8+4 \cos 60^{\circ}-\mathrm{F}_{2} \cos 30^{\circ}=0\) \(8+2-\mathrm{F}_{2} \frac{\sqrt{3}}{2}=0\) \(\mathrm{~F}_{2}=\frac{20}{\sqrt{3}} \mathrm{~N}\) Now, (ii) For the force acting in along the y-axis \(\left(\sum \mathrm{F}_{\mathrm{y}}=0\right)\) \(\therefore \mathrm{F}_{1}+4 \sin 60^{\circ}-\mathrm{F}_{2} \sin 30^{\circ}=0\) \(\mathrm{~F}_{1}+\frac{4 \sqrt{3}}{2}-\frac{\mathrm{F}_{2}}{2}=0\) \(\mathrm{~F}_{1}=\frac{\mathrm{F}_{2}}{2}-2 \sqrt{3}=\frac{10}{\sqrt{3}}-2 \sqrt{3}\) \(\mathrm{~F}_{1}=\frac{4}{\sqrt{3}} \mathrm{~N}\) Hence, magnitude of \(\mathrm{F}_{1}=\frac{4}{\sqrt{3}} \mathrm{~N}\) and \(\text { Magnitude of } F_{2}=\frac{20}{\sqrt{3}} \mathrm{~N}\)
