Forces of 10.6 N north, 20.8 N east, and 14.3 N south are simultaneously applied to a 3.76 kg mass as it rests on an air table. What is the magnitude of its acceleration? What is the direction of the acceleration in degrees? (Take east to be 0 degrees and counterclockwise to be positive.)
I am tired of answering this. I have given you the angle S of East. Convert it to your system.
I will be happy to critique your work.
What do you mean by convert it to your system?
I gave you degrees S of E. You are told to have E zero, and measure the angle counterclockwise. I did it clockwise.
So tan theta = 20.8/3.7?
Yes, that is theta measured clockwise from East. That is not the way your problem stated to give the answer.
i tried both ways and both are wrong :s
To find the magnitude of acceleration, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, this is expressed as:
ΣF = m*a,
where ΣF is the net force, m is the mass, and a is the acceleration.
In this case, we have three forces acting on the object: 10.6 N north, 20.8 N east, and 14.3 N south. To find the net force, we need to add these forces together, taking into account their directions.
Let's break down the forces into their x (east/west) and y (north/south) components.
The 20.8 N force in the east direction does not have a north/south component, so its x component is 20.8 N and its y component is 0 N.
The 10.6 N force in the north direction does not have an east/west component, so its x component is 0 N and its y component is 10.6 N.
The 14.3 N force in the south direction does not have an east/west component, so its x component is 0 N and its y component is -14.3 N.
Now, let's add the x and y components separately to find the net force:
Net x force = 20.8 N + 0 N + 0 N = 20.8 N
Net y force = 0 N + 10.6 N - 14.3 N = -3.7 N
Now we can find the magnitude of the net force using the Pythagorean theorem:
ΣF = √((Net x force)^2 + (Net y force)^2)
= √((20.8 N)^2 + (-3.7 N)^2)
= √(432.64 N^2 + 13.69 N^2)
= √446.33 N^2
≈ 21.13 N
The magnitude of the net force is approximately 21.13 N.
To find the direction of the acceleration in degrees, we can use trigonometry. The direction can be expressed as an angle measured counterclockwise from the east direction, which we have taken as 0 degrees.
The tangent of the angle can be found as follows:
tan(θ) = (Net y force) / (Net x force)
= (-3.7 N) / (20.8 N)
≈ -0.178
To find the angle, we take the arctangent of the tangent:
θ = arctan(-0.178)
≈ -9.85 degrees
Since we want a positive angle, we can add 360 degrees to the negative angle:
θ = -9.85 + 360
≈ 350.15 degrees
The direction of the acceleration is approximately 350.15 degrees counterclockwise from the east direction.