A 15N force acting at 15 degrees north of east and a 18N force acting at 79 degrees north of west act concurrently on an object. What is the magnitude and direction of a third force that produces equilibrium on the object? Show sketch and work.
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x-component:: 15*cos(15 deg)+18cos(101 deg) = 11.05
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y-component:: 15*sin(15 deg)+18sin(101 deg) = 21.55
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magnitude = sqrt[11.05^2+21.55^2] = 24.22
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direction = arctan(21.55/11.05) = 62.85 degrees north of east
That is the resultant, but you need the opposite to that so move it around 180 to cancel it
To find the magnitude and direction of the third force that produces equilibrium on the object, we need to add the two given forces together. We can break down each force into its x and y components.
For the 15N force acting at 15 degrees north of east:
x-component: 15 * cos(15 degrees) = 15 * 0.9659 = 14.49N acting to the east
y-component: 15 * sin(15 degrees) = 15 * 0.2588 = 3.88N acting to the north
For the 18N force acting at 79 degrees north of west:
x-component: 18 * cos(79 degrees) = 18 * (-0.1450) = -2.61N acting to the east (negative because it's in the opposite direction)
y-component: 18 * sin(79 degrees) = 18 * 0.9894 = 17.81N acting to the north
To find the x-component of the third force, we add the x-components of the two forces together:
x-component = 14.49N + (-2.61N) = 11.88N acting to the east
To find the y-component of the third force, we add the y-components of the two forces together:
y-component = 3.88N + 17.81N = 21.69N acting to the north
Now, we can use the x and y components to find the magnitude and direction of the third force.
The magnitude of the third force, also known as the resultant force, can be found using the Pythagorean theorem:
magnitude = √(x-component^2 + y-component^2)
magnitude = √(11.88N^2 + 21.69N^2)
magnitude = √(140.9944N^2 + 470.9161N^2)
magnitude = √611.9105N^2
magnitude ≈ 24.22N (rounded to two decimal places)
The direction of the third force can be found using trigonometry. We need to find the angle that the resultant force makes with the east direction (positive x-axis). We can use the inverse tangent function (arctan) to find this angle:
direction = arctan(y-component / x-component)
direction = arctan(21.69N / 11.88N)
direction ≈ 62.85 degrees north of east (rounded to two decimal places)
To summarize, the magnitude of the third force is approximately 24.22N, and its direction is approximately 62.85 degrees north of east.