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Sagot :
Sure, let's break down the steps to solve this problem. We have a string fastened to an object weighing [tex]\( 8 \text{ units} \)[/tex]. When pulled by a horizontal force [tex]\( P \)[/tex], the string makes an angle of [tex]\( 30^{\circ} \)[/tex] with the vertical wall.
a) Tension in the string ( [tex]\( T \)[/tex] ):
1. Identify the vertical and horizontal components:
- The weight of the object acts vertically downward and has a magnitude of [tex]\( 8 \text{ units} \)[/tex].
- The tension [tex]\( T \)[/tex] in the string has both vertical and horizontal components because the string is inclined.
2. Determine the vertical component of the tension:
Since the string makes an angle of [tex]\( 30^{\circ} \)[/tex] with the vertical, the vertical component of the tension in the string is given by [tex]\( T \cos(30^{\circ}) \)[/tex].
The vertical component of this tension must balance the weight of the object:
[tex]\[ T \cos(30^{\circ}) = 8 \][/tex]
3. Solve for the tension [tex]\( T \)[/tex]:
Knowing that [tex]\( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \)[/tex], we can rewrite the equation:
[tex]\[ T \frac{\sqrt{3}}{2} = 8 \][/tex]
Solving for [tex]\( T \)[/tex], we get:
[tex]\[ T = \frac{8}{\frac{\sqrt{3}}{2}} = \frac{8 \times 2}{\sqrt{3}} = \frac{16}{\sqrt{3}} \][/tex]
4. Simplify and round off the result:
Dividing and rationalizing the denominator is not necessary here, but rounding to two decimal places:
[tex]\[ T \approx 9.24 \][/tex]
So, the tension in the string, [tex]\( T \)[/tex], is approximately [tex]\( 9.24 \)[/tex] units.
b) Horizontal force ( [tex]\( P \)[/tex] ):
1. Determine the horizontal component of the tension:
The horizontal component of [tex]\( T \)[/tex] is given by [tex]\( T \sin(30^{\circ}) \)[/tex].
2. Solve for the horizontal force [tex]\( P \)[/tex]:
Since [tex]\( \sin(30^{\circ}) = \frac{1}{2} \)[/tex],
[tex]\[ P = T \sin(30^{\circ}) = T \times \frac{1}{2} \][/tex]
3. Substitute the value of [tex]\( T \)[/tex]:
Using [tex]\( T \approx 9.24 \)[/tex], we find:
[tex]\[ P \approx 9.24 \times \frac{1}{2} = 4.62 \][/tex]
So, the horizontal force, [tex]\( P \)[/tex], is approximately [tex]\( 4.62 \)[/tex] units.
Summary:
- Tension in the string ( [tex]\( T \)[/tex] ): [tex]\( 9.24 \)[/tex]
- Horizontal force ( [tex]\( P \)[/tex] ): [tex]\( 4.62 \)[/tex]
Both values are correct to two decimal places.
a) Tension in the string ( [tex]\( T \)[/tex] ):
1. Identify the vertical and horizontal components:
- The weight of the object acts vertically downward and has a magnitude of [tex]\( 8 \text{ units} \)[/tex].
- The tension [tex]\( T \)[/tex] in the string has both vertical and horizontal components because the string is inclined.
2. Determine the vertical component of the tension:
Since the string makes an angle of [tex]\( 30^{\circ} \)[/tex] with the vertical, the vertical component of the tension in the string is given by [tex]\( T \cos(30^{\circ}) \)[/tex].
The vertical component of this tension must balance the weight of the object:
[tex]\[ T \cos(30^{\circ}) = 8 \][/tex]
3. Solve for the tension [tex]\( T \)[/tex]:
Knowing that [tex]\( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \)[/tex], we can rewrite the equation:
[tex]\[ T \frac{\sqrt{3}}{2} = 8 \][/tex]
Solving for [tex]\( T \)[/tex], we get:
[tex]\[ T = \frac{8}{\frac{\sqrt{3}}{2}} = \frac{8 \times 2}{\sqrt{3}} = \frac{16}{\sqrt{3}} \][/tex]
4. Simplify and round off the result:
Dividing and rationalizing the denominator is not necessary here, but rounding to two decimal places:
[tex]\[ T \approx 9.24 \][/tex]
So, the tension in the string, [tex]\( T \)[/tex], is approximately [tex]\( 9.24 \)[/tex] units.
b) Horizontal force ( [tex]\( P \)[/tex] ):
1. Determine the horizontal component of the tension:
The horizontal component of [tex]\( T \)[/tex] is given by [tex]\( T \sin(30^{\circ}) \)[/tex].
2. Solve for the horizontal force [tex]\( P \)[/tex]:
Since [tex]\( \sin(30^{\circ}) = \frac{1}{2} \)[/tex],
[tex]\[ P = T \sin(30^{\circ}) = T \times \frac{1}{2} \][/tex]
3. Substitute the value of [tex]\( T \)[/tex]:
Using [tex]\( T \approx 9.24 \)[/tex], we find:
[tex]\[ P \approx 9.24 \times \frac{1}{2} = 4.62 \][/tex]
So, the horizontal force, [tex]\( P \)[/tex], is approximately [tex]\( 4.62 \)[/tex] units.
Summary:
- Tension in the string ( [tex]\( T \)[/tex] ): [tex]\( 9.24 \)[/tex]
- Horizontal force ( [tex]\( P \)[/tex] ): [tex]\( 4.62 \)[/tex]
Both values are correct to two decimal places.
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