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Laws of Motion

146002 The resultant of two forces acting at an angle of $120^{\circ}$ is $10 kg$ -wt and is perpendicular to one of the forces. That force is:

1

10 \sqrt{3} kg - wt

2

20 \sqrt{3} kg - wt

3

10 kg - wt

4

\frac{10}{\sqrt{3}} kg - wt

Explanation:

D Given,
$F = 10 kg - wt$
$θ = 120^{\circ}$
From fig (a)

Here,
$A$ and $B$ are two forces acting from the same point and $C$ is their resultant,
So, the angle $b / w$ the two forces $A$ and $B$ is $120^{\circ}$ .

From figure (b)
In $△ ACD$
$\tan θ = \frac{force that we need to find}{resultant force} = \frac{AC}{CD}$
$\tan 30^{\circ} = \frac{F_{1}}{F}$
$F_{1} = F \tan 30^{\circ}$
$F_{1} = 10 \times \frac{1}{\sqrt{3}} kg - wt$

Karnataka CET-2011

Laws of Motion

146003 A steel wire can withstand a load up to $2940 N$ . A load of $150 kg$ is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is

1

30^{\circ}

2

60^{\circ}

3

80^{\circ}

4

85^{\circ}

Explanation:

B Given,
Load (T) $= 2940 N$
Mass $(m) = 150 kg$
$g = 9.8 m / s^{2}$
$θ =$ ?

(I)
(II)
Initial condition Maximum angle conditon At point B in (ii) Figure in equilibrium condition
$T \cos θ = mg$
$\cos θ = \frac{mg}{T}$
$\cos θ = \frac{150 \times 9.8}{2940}$
$\cos θ = 0.5 = \frac{1}{2}$
$\cos θ = \cos 60^{\circ}$
$\Rightarrow θ = 60^{\circ}$
Hence, the maximum angle $θ = 60^{\circ}$ at which the wire can be displaced from the mean position.

EAMCET-2008

Laws of Motion

146005 Forces of $5 N, 12 N$ and $13 N$ are in equilibrium. If $\sin 23^{\circ} = \frac{5}{13}$ , then the angle
between $5 N$ and $13 N$ forces is

1

23^{\circ}

2

67^{\circ}

3

90^{\circ}

4

113^{\circ}

Explanation:

B Given,
Forces $5 N, 12 N, 13 N$ are in Equilibrium.
Given, $\sin 23^{\circ} = \frac{5}{13}$

We know, in $△ ABC$
$∠ A + ∠ B + ∠ C = 180^{\circ}$
$α + 90^{\circ} + 23^{\circ} = 180^{\circ}$
$α = 180^{\circ} - 113^{\circ}$
$α = 67^{\circ}$

Assam CEE-2014

Laws of Motion

146006 The sum of magnitudes of two forces acting at a point is $16 N$ . If their resultant is normal to smaller force, and has a magnitude $8 N$ , then forces are

1

6 N, 10 N

2

8 N, 8 N

3

4 N, 12 N

4

2 N, 14 N

Explanation:

A Let a and $b$ is two forces.
Then given,
$| \vec{a} | + | \vec{b} | = 16 N, | \vec{R} | = 8 N$
$∴ \vec{R} = \vec{a} + \vec{b}$
Squaring both side,
$(\vec{R})^{2} = (\vec{a} + \vec{b})^{2}$
$| \vec{R} | = \sqrt{a^{2} + b^{2} + 2 ab \cos θ}$
$Let, \vec{a} > \vec{b}$
$\vec{R} \cdot \vec{b} = 0$
$(\vec{a} + \vec{b}) \cdot \vec{b} = 0$
$\vec{a} \cdot \vec{b} = - (b)^{2}$
$ab \cos ◻ = - b^{2}$
$8 = \sqrt{(16 - b)^{2} + b^{2} + 2 (- b)^{2}}$
Squaring both side-
$64 = 256 + b^{2} - 32 b + b^{2} - 2 b^{2}$
$32 b = 256 - 64$
$32 b = 192$
$b = 6 N$
$∴ a = 16 - 6 = 10 N$

AP EAMCET -2012

Laws of Motion

146002 The resultant of two forces acting at an angle of $120^{\circ}$ is $10 kg$ -wt and is perpendicular to one of the forces. That force is:

1

10 \sqrt{3} kg - wt

2

20 \sqrt{3} kg - wt

3

10 kg - wt

4

\frac{10}{\sqrt{3}} kg - wt

Explanation:

D Given,
$F = 10 kg - wt$
$θ = 120^{\circ}$
From fig (a)

Here,
$A$ and $B$ are two forces acting from the same point and $C$ is their resultant,
So, the angle $b / w$ the two forces $A$ and $B$ is $120^{\circ}$ .

From figure (b)
In $△ ACD$
$\tan θ = \frac{force that we need to find}{resultant force} = \frac{AC}{CD}$
$\tan 30^{\circ} = \frac{F_{1}}{F}$
$F_{1} = F \tan 30^{\circ}$
$F_{1} = 10 \times \frac{1}{\sqrt{3}} kg - wt$

Karnataka CET-2011

Laws of Motion

146003 A steel wire can withstand a load up to $2940 N$ . A load of $150 kg$ is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is

1

30^{\circ}

2

60^{\circ}

3

80^{\circ}

4

85^{\circ}

Explanation:

B Given,
Load (T) $= 2940 N$
Mass $(m) = 150 kg$
$g = 9.8 m / s^{2}$
$θ =$ ?

(I)
(II)
Initial condition Maximum angle conditon At point B in (ii) Figure in equilibrium condition
$T \cos θ = mg$
$\cos θ = \frac{mg}{T}$
$\cos θ = \frac{150 \times 9.8}{2940}$
$\cos θ = 0.5 = \frac{1}{2}$
$\cos θ = \cos 60^{\circ}$
$\Rightarrow θ = 60^{\circ}$
Hence, the maximum angle $θ = 60^{\circ}$ at which the wire can be displaced from the mean position.

EAMCET-2008

Laws of Motion

146005 Forces of $5 N, 12 N$ and $13 N$ are in equilibrium. If $\sin 23^{\circ} = \frac{5}{13}$ , then the angle
between $5 N$ and $13 N$ forces is

1

23^{\circ}

2

67^{\circ}

3

90^{\circ}

4

113^{\circ}

Explanation:

B Given,
Forces $5 N, 12 N, 13 N$ are in Equilibrium.
Given, $\sin 23^{\circ} = \frac{5}{13}$

We know, in $△ ABC$
$∠ A + ∠ B + ∠ C = 180^{\circ}$
$α + 90^{\circ} + 23^{\circ} = 180^{\circ}$
$α = 180^{\circ} - 113^{\circ}$
$α = 67^{\circ}$

Assam CEE-2014

Laws of Motion

146006 The sum of magnitudes of two forces acting at a point is $16 N$ . If their resultant is normal to smaller force, and has a magnitude $8 N$ , then forces are

1

6 N, 10 N

2

8 N, 8 N

3

4 N, 12 N

4

2 N, 14 N

Explanation:

A Let a and $b$ is two forces.
Then given,
$| \vec{a} | + | \vec{b} | = 16 N, | \vec{R} | = 8 N$
$∴ \vec{R} = \vec{a} + \vec{b}$
Squaring both side,
$(\vec{R})^{2} = (\vec{a} + \vec{b})^{2}$
$| \vec{R} | = \sqrt{a^{2} + b^{2} + 2 ab \cos θ}$
$Let, \vec{a} > \vec{b}$
$\vec{R} \cdot \vec{b} = 0$
$(\vec{a} + \vec{b}) \cdot \vec{b} = 0$
$\vec{a} \cdot \vec{b} = - (b)^{2}$
$ab \cos ◻ = - b^{2}$
$8 = \sqrt{(16 - b)^{2} + b^{2} + 2 (- b)^{2}}$
Squaring both side-
$64 = 256 + b^{2} - 32 b + b^{2} - 2 b^{2}$
$32 b = 256 - 64$
$32 b = 192$
$b = 6 N$
$∴ a = 16 - 6 = 10 N$

AP EAMCET -2012

Laws of Motion

146002 The resultant of two forces acting at an angle of $120^{\circ}$ is $10 kg$ -wt and is perpendicular to one of the forces. That force is:

1

10 \sqrt{3} kg - wt

2

20 \sqrt{3} kg - wt

3

10 kg - wt

4

\frac{10}{\sqrt{3}} kg - wt

Explanation:

D Given,
$F = 10 kg - wt$
$θ = 120^{\circ}$
From fig (a)

Here,
$A$ and $B$ are two forces acting from the same point and $C$ is their resultant,
So, the angle $b / w$ the two forces $A$ and $B$ is $120^{\circ}$ .

From figure (b)
In $△ ACD$
$\tan θ = \frac{force that we need to find}{resultant force} = \frac{AC}{CD}$
$\tan 30^{\circ} = \frac{F_{1}}{F}$
$F_{1} = F \tan 30^{\circ}$
$F_{1} = 10 \times \frac{1}{\sqrt{3}} kg - wt$

Karnataka CET-2011

Laws of Motion

146003 A steel wire can withstand a load up to $2940 N$ . A load of $150 kg$ is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is

1

30^{\circ}

2

60^{\circ}

3

80^{\circ}

4

85^{\circ}

Explanation:

B Given,
Load (T) $= 2940 N$
Mass $(m) = 150 kg$
$g = 9.8 m / s^{2}$
$θ =$ ?

(I)
(II)
Initial condition Maximum angle conditon At point B in (ii) Figure in equilibrium condition
$T \cos θ = mg$
$\cos θ = \frac{mg}{T}$
$\cos θ = \frac{150 \times 9.8}{2940}$
$\cos θ = 0.5 = \frac{1}{2}$
$\cos θ = \cos 60^{\circ}$
$\Rightarrow θ = 60^{\circ}$
Hence, the maximum angle $θ = 60^{\circ}$ at which the wire can be displaced from the mean position.

EAMCET-2008

Laws of Motion

146005 Forces of $5 N, 12 N$ and $13 N$ are in equilibrium. If $\sin 23^{\circ} = \frac{5}{13}$ , then the angle
between $5 N$ and $13 N$ forces is

1

23^{\circ}

2

67^{\circ}

3

90^{\circ}

4

113^{\circ}

Explanation:

B Given,
Forces $5 N, 12 N, 13 N$ are in Equilibrium.
Given, $\sin 23^{\circ} = \frac{5}{13}$

We know, in $△ ABC$
$∠ A + ∠ B + ∠ C = 180^{\circ}$
$α + 90^{\circ} + 23^{\circ} = 180^{\circ}$
$α = 180^{\circ} - 113^{\circ}$
$α = 67^{\circ}$

Assam CEE-2014

Laws of Motion

146006 The sum of magnitudes of two forces acting at a point is $16 N$ . If their resultant is normal to smaller force, and has a magnitude $8 N$ , then forces are

1

6 N, 10 N

2

8 N, 8 N

3

4 N, 12 N

4

2 N, 14 N

Explanation:

A Let a and $b$ is two forces.
Then given,
$| \vec{a} | + | \vec{b} | = 16 N, | \vec{R} | = 8 N$
$∴ \vec{R} = \vec{a} + \vec{b}$
Squaring both side,
$(\vec{R})^{2} = (\vec{a} + \vec{b})^{2}$
$| \vec{R} | = \sqrt{a^{2} + b^{2} + 2 ab \cos θ}$
$Let, \vec{a} > \vec{b}$
$\vec{R} \cdot \vec{b} = 0$
$(\vec{a} + \vec{b}) \cdot \vec{b} = 0$
$\vec{a} \cdot \vec{b} = - (b)^{2}$
$ab \cos ◻ = - b^{2}$
$8 = \sqrt{(16 - b)^{2} + b^{2} + 2 (- b)^{2}}$
Squaring both side-
$64 = 256 + b^{2} - 32 b + b^{2} - 2 b^{2}$
$32 b = 256 - 64$
$32 b = 192$
$b = 6 N$
$∴ a = 16 - 6 = 10 N$

AP EAMCET -2012

Laws of Motion

146002 The resultant of two forces acting at an angle of $120^{\circ}$ is $10 kg$ -wt and is perpendicular to one of the forces. That force is:

1

10 \sqrt{3} kg - wt

2

20 \sqrt{3} kg - wt

3

10 kg - wt

4

\frac{10}{\sqrt{3}} kg - wt

Explanation:

D Given,
$F = 10 kg - wt$
$θ = 120^{\circ}$
From fig (a)

Here,
$A$ and $B$ are two forces acting from the same point and $C$ is their resultant,
So, the angle $b / w$ the two forces $A$ and $B$ is $120^{\circ}$ .

From figure (b)
In $△ ACD$
$\tan θ = \frac{force that we need to find}{resultant force} = \frac{AC}{CD}$
$\tan 30^{\circ} = \frac{F_{1}}{F}$
$F_{1} = F \tan 30^{\circ}$
$F_{1} = 10 \times \frac{1}{\sqrt{3}} kg - wt$

Karnataka CET-2011

Laws of Motion

146003 A steel wire can withstand a load up to $2940 N$ . A load of $150 kg$ is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is

1

30^{\circ}

2

60^{\circ}

3

80^{\circ}

4

85^{\circ}

Explanation:

B Given,
Load (T) $= 2940 N$
Mass $(m) = 150 kg$
$g = 9.8 m / s^{2}$
$θ =$ ?

(I)
(II)
Initial condition Maximum angle conditon At point B in (ii) Figure in equilibrium condition
$T \cos θ = mg$
$\cos θ = \frac{mg}{T}$
$\cos θ = \frac{150 \times 9.8}{2940}$
$\cos θ = 0.5 = \frac{1}{2}$
$\cos θ = \cos 60^{\circ}$
$\Rightarrow θ = 60^{\circ}$
Hence, the maximum angle $θ = 60^{\circ}$ at which the wire can be displaced from the mean position.

EAMCET-2008

Laws of Motion

146005 Forces of $5 N, 12 N$ and $13 N$ are in equilibrium. If $\sin 23^{\circ} = \frac{5}{13}$ , then the angle
between $5 N$ and $13 N$ forces is

1

23^{\circ}

2

67^{\circ}

3

90^{\circ}

4

113^{\circ}

Explanation:

B Given,
Forces $5 N, 12 N, 13 N$ are in Equilibrium.
Given, $\sin 23^{\circ} = \frac{5}{13}$

We know, in $△ ABC$
$∠ A + ∠ B + ∠ C = 180^{\circ}$
$α + 90^{\circ} + 23^{\circ} = 180^{\circ}$
$α = 180^{\circ} - 113^{\circ}$
$α = 67^{\circ}$

Assam CEE-2014

Laws of Motion

146006 The sum of magnitudes of two forces acting at a point is $16 N$ . If their resultant is normal to smaller force, and has a magnitude $8 N$ , then forces are

1

6 N, 10 N

2

8 N, 8 N

3

4 N, 12 N

4

2 N, 14 N

Explanation:

A Let a and $b$ is two forces.
Then given,
$| \vec{a} | + | \vec{b} | = 16 N, | \vec{R} | = 8 N$
$∴ \vec{R} = \vec{a} + \vec{b}$
Squaring both side,
$(\vec{R})^{2} = (\vec{a} + \vec{b})^{2}$
$| \vec{R} | = \sqrt{a^{2} + b^{2} + 2 ab \cos θ}$
$Let, \vec{a} > \vec{b}$
$\vec{R} \cdot \vec{b} = 0$
$(\vec{a} + \vec{b}) \cdot \vec{b} = 0$
$\vec{a} \cdot \vec{b} = - (b)^{2}$
$ab \cos ◻ = - b^{2}$
$8 = \sqrt{(16 - b)^{2} + b^{2} + 2 (- b)^{2}}$
Squaring both side-
$64 = 256 + b^{2} - 32 b + b^{2} - 2 b^{2}$
$32 b = 256 - 64$
$32 b = 192$
$b = 6 N$
$∴ a = 16 - 6 = 10 N$

AP EAMCET -2012

Question Bank Series

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