In a physics laboratory class, three massless ropes are tied together at a point. A pulling force is applied along each rope: \(F_{1}=150 . \mathrm{N}\) at \(60.0^{\circ}, F_{2}=200 . \mathrm{N}\) at \(100 .^{\circ}, F_{3}=100 . \mathrm{N}\) at \(190 .^{\circ} .\) What is the magnitude of a fourth force and the angle at which it acts to keep the point at the center of the system stationary? (All angles are measured from the positive \(x\) -axis.)

Short Answer

Expert verified
Short Answer: To find the magnitude and direction of the fourth force needed to keep the system stationary, follow these steps: 1. Express each given force as a vector with components in x and y directions using trigonometry. 2. Find the total x and y components of the forces acting on the system by summing the individual components. 3. Calculate the x and y components of the fourth force required to make the total force in both x and y directions zero. 4. Find the magnitude and direction of the fourth force using the calculated components and the arctan function. 5. Convert the direction to the appropriate angle with respect to the positive x-axis, ensuring it's consistent with the calculated components of the fourth force.

Step by step solution

01

Express each force as a vector with components in x and y directions.

To find the components of \(\vec{F}_{1}\), we use basic trigonometry: - \(F_{1x} = F_{1} \cos{60^{\circ}} = 150 \cos{60^{\circ}}\) - \(F_{1y} = F_{1} \sin{60^{\circ}} = 150 \sin{60^{\circ}}\) For \(\vec{F}_{2}\): - \(F_{2x} = F_{2} \cos{100^{\circ}} = 200 \cos{100^{\circ}}\) - \(F_{2y} = F_{2} \sin{100^{\circ}} = 200 \sin{100^{\circ}}\) For \(\vec{F}_{3}\): - \(F_{3x} = F_{3} \cos{190^{\circ}} = 100 \cos{190^{\circ}}\) - \(F_{3y} = F_{3} \sin{190^{\circ}} = 100 \sin{190^{\circ}}\)
02

Find the total x and y components of the forces acting on the system.

Now that we have the components of each force, we will sum the components in the x and y directions. - \(F_{x(total)} = F_{1x} + F_{2x} + F_{3x}\) - \(F_{y(total)} = F_{1y} + F_{2y} + F_{3y}\)
03

Calculate the fourth force needed to keep the system stationary.

Since the system is to be kept stationary, the total force in x and y directions should be zero. - \(F_{4x} = -F_{x(total)}\) - \(F_{4y} = -F_{y(total)}\)
04

Calculate the magnitude and direction of the fourth force.

Now that we have the x and y components of the fourth force, the magnitude and the angle can be calculated by: - Magnitude: \(F_{4} = \sqrt{F_{4x}^2 + F_{4y}^2}\) - Direction: \(\theta_{4} = \arctan{\frac{F_{4y}}{F_{4x}}}\)
05

Convert the direction to the appropriate angle with respect to the positive x-axis.

Since the angle reported by the arctan function can be ambiguous, make sure to verify and correct \(\theta_{4}\) if needed, to ensure the angle is consistent with the direction of the calculated components of \(\vec{F}_{4}\). The magnitude and direction of the force \(\vec{F}_4\) will keep the system stationary.

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