A body at equilibrium is under the action of three forces. One force is 10N acting due east and one is 5N in the direction 60 degree north-east. What is the magnitude and direction of the third force?
F(1x)=F(1)=10 N
F(2x)=F(2)cos60=5•0.5=2.5 N
F(2y) =F(2)sin60 = 5•0.866= 4.33 N
F(3x) =- F(x)=- (10+2.5 )= -12.5 N
F(3y) =- F(2y)= - 4.33 N
F(3) = sqrt{ F(3x)²+F(3y)²} =13.23 N
tan φ = F(3y)/F(3x) =4.33/12.5=0.364
φ = 19.1⁰ (south-west)
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Well, it seems like this body is having a party with all these forces! Let's figure out the magnitude and direction of the third force together.
If we break down the 5N force in the north-east direction into its components, we can figure out its effect on the equilibrium. The north component would be 5N * cos(60 degrees), and the east component would be 5N * sin(60 degrees).
Now, let's add up the eastward force component from the 10N force together with the eastward component of the 5N force. The total eastward force would be 10N + (5N * sin(60 degrees)).
Similarly, we add up the northward force component from the 10N force together with the northward component of the 5N force. The total northward force would be 5N * cos(60 degrees) + (10N * sin(60 degrees)).
To find the magnitude of the third force, we can use the Pythagorean theorem: magnitude = √(eastward force^2 + northward force^2).
As for the direction, we can use a bit of trigonometry. We can take the inverse tangent of the eastward force divided by the northward force to find the angle.
But, beware, my friend! Remember that the direction may be given in terms of reference angles, such as from the positive x-axis or positive y-axis. So, make sure to adjust the angle accordingly.
But hey, don't worry too much! With a little math magic, you'll figure it out in no time!
To find the magnitude and direction of the third force, we can use vector addition. Let's break down the given forces into their x and y components.
The first force of 10N acting due east has an x-component of 10N and a y-component of 0N since it is only acting in the east direction.
The second force of 5N in the direction 60 degrees north-east can be split into its x and y components using trigonometry. The angle of 60 degrees is measured from the positive x-axis.
The x-component can be found using the cosine function: cos(60°) = adjacent/hypotenuse = x-component/5N.
Solving for the x-component: x-component = 5N * cos(60°) = 5N * 0.5 = 2.5N.
The y-component can be found using the sine function: sin(60°) = opposite/hypotenuse = y-component/5N.
Solving for the y-component: y-component = 5N * sin(60°) = 5N * (√3/2) ≈ 4.33N.
Now, we can find the resulting x and y components by adding the x and y components of the two forces.
The x-component of the resulting force is the sum of the x-components of the two forces:
Resultant x-component = 10N + 2.5N = 12.5N.
The y-component of the resulting force is the sum of the y-components of the two forces:
Resultant y-component = 0N + 4.33N = 4.33N.
To find the magnitude of the resulting force, we use the Pythagorean theorem:
Magnitude of the resulting force = √(Resultant x-component)^2 + (Resultant y-component)^2
= √(12.5N)^2 + (4.33N)^2
= √(156.25N^2) + (18.7489N^2)
= √(175.9989N^2)
≈ 13.26N.
To find the direction of the resulting force, we can use trigonometry again. The angle θ can be found using the inverse tangent (arctan) function:
θ = arctan(Resultant y-component/Resultant x-component)
≈ arctan(4.33N/12.5N)
≈ 19.13°.
Therefore, the magnitude of the third force is approximately 13.26N, and the direction is approximately 19.13 degrees with respect to the east direction.