Question

A wind blows on a m=4 kg mass that hangs from a 2 m long string by exerting a constant horizontal force of F=5N.
What is ils equilibrium height, h?
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Answer 1

Given:

A wind blows on a m=4 kg mass that hangs from a 2 m long string by exerting a constant horizontal force of F=5N.

To find:

Equilibrium height?

Calculation:

After dividing the components, we can say:

$1) \: T \sin( \theta) = F$

$2) \: T \cos( \theta) = mg$

Dividing the Equations:

$\implies \: \tan( \theta) = \dfrac{F}{mg}$

$\implies \: \tan( \theta) = \dfrac{5}{40}$

$\implies \: \tan( \theta) = \dfrac{1}{8}$

Now, we can say:

$\cos( \theta) = \dfrac{8}{ \sqrt{65} }$

$\implies \: \dfrac{x}{r} = \dfrac{8}{ \sqrt{65} }$

$\implies \: \dfrac{x}{2} = \dfrac{8}{ \sqrt{65} }$

$\implies \: x = \dfrac{16}{ \sqrt{65} }$

$\implies \: x = 1.98 \: m$

So, equilibrium height will be :

$\implies \: h = 2 - 1.98 = 0.02m$

So, equilibrium height is 0.02 metres.

Answer 2

Explanation:

Given:

A wind blows on a m=4 kg mass that hangs from a 2 m long string by exerting a constant horizontal force of F=5N.

To find:

Equilibrium height?

Calculation:

After dividing the components, we can say:

1) \: T \sin( \theta) = F1)Tsin(θ)=F

2) \: T \cos( \theta) = mg2)Tcos(θ)=mg

Dividing the Equations:

\implies \: \tan( \theta) = \dfrac{F}{mg} ⟹tan(θ)=

mg

F

\implies \: \tan( \theta) = \dfrac{5}{40} ⟹tan(θ)=

40

5

\implies \: \tan( \theta) = \dfrac{1}{8} ⟹tan(θ)=

8

1

Now, we can say:

\cos( \theta) = \dfrac{8}{ \sqrt{65} } cos(θ)=

65

8

\implies \: \dfrac{x}{r} = \dfrac{8}{ \sqrt{65} } ⟹

r

x

=

65

8

\implies \: \dfrac{x}{2} = \dfrac{8}{ \sqrt{65} } ⟹

2

x

=

65

8

\implies \: x = \dfrac{16}{ \sqrt{65} } ⟹x=

65

16

\implies \: x = 1.98 \: m⟹x=1.98m

So, equilibrium height will be :

\implies \: h = 2 - 1.98 = 0.02m⟹h=2−1.98=0.02m

So, equilibrium height is 0.02 metres.