Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 68.0° counter-clockwise from the x axis, while the 2nd force has x- and y-components (-19.0 N, 16.5 N). Express the magnitude and direction of the net force.
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F(x) =17cos68-19 =
F(y) = 17sin68+16.5 =
F=sqrt{F(x)²+F(y)²}=
tanα =F(y)/F(x)=
To find the magnitude and direction of the net force, we can break down each force into its x and y components and then add them together.
Given:
Force 1 magnitude = 17.0 N
Force 1 angle = 68.0° counter-clockwise from the x-axis
Force 2 x-component = -19.0 N
Force 2 y-component = 16.5 N
To find the x and y components of Force 1:
Force 1 x-component = Force 1 magnitude * cos(angle)
Force 1 x-component = 17.0 N * cos(68.0°)
Force 1 x-component = 17.0 N * 0.389
Force 1 x-component = 6.60 N
Force 1 y-component = Force 1 magnitude * sin(angle)
Force 1 y-component = 17.0 N * sin(68.0°)
Force 1 y-component = 17.0 N * 0.920
Force 1 y-component = 15.64 N
To find the net x and y components of the forces:
Net x-component = Force 1 x-component + Force 2 x-component
Net x-component = 6.60 N + (-19.0 N)
Net x-component = -12.4 N
Net y-component = Force 1 y-component + Force 2 y-component
Net y-component = 15.64 N + 16.5 N
Net y-component = 32.14 N
Now, we can find the magnitude and direction of the net force:
Magnitude of the net force = √(Net x-component^2 + Net y-component^2)
Magnitude of the net force = √((-12.4 N)^2 + (32.14 N)^2)
Magnitude of the net force = √(153.76 N^2 + 1033.44 N^2)
Magnitude of the net force = √(1187.20 N^2)
Magnitude of the net force ≈ 34.45 N
To find the direction of the net force:
Direction of the net force = arctan(Net y-component / Net x-component)
Direction of the net force = arctan(32.14 N / -12.4 N)
Direction of the net force ≈ -69.7°
Therefore, the magnitude of the net force is approximately 34.45 N and its direction is approximately 69.7° counter-clockwise from the x-axis.
To find the net force, we need to add the two forces vectorially.
First, let's break down the first force into its x and y components.
Given that the magnitude of the first force is 17.0 N and the angle it makes with the x-axis is 68.0° counter-clockwise, we can use trigonometry to find its components.
The x-component (Fx) can be calculated using the formula Fx = F * cos(θ), where F is the magnitude of the force and θ is the angle it makes with the x-axis.
Fx = 17.0 N * cos(68.0°)
Fx = 17.0 N * 0.389
Fx = 6.61 N
The y-component (Fy) can be calculated using the formula Fy = F * sin(θ), where F is the magnitude of the force and θ is the angle it makes with the x-axis.
Fy = 17.0 N * sin(68.0°)
Fy = 17.0 N * 0.921
Fy = 15.66 N
For the second force, the x-component is given as -19.0 N and the y-component is given as 16.5 N.
Next, we can find the net x and y components by adding the corresponding components of the two forces.
Net Fx = Fx1 + Fx2 = 6.61 N + (-19.0 N) = -12.39 N
Net Fy = Fy1 + Fy2 = 15.66 N + 16.5 N = 32.16 N
Finally, we can find the magnitude of the net force using the Pythagorean theorem.
Magnitude of the net force (Fnet) = √(Net Fx^2 + Net Fy^2)
Fnet = √((-12.39 N)^2 + (32.16 N)^2)
Fnet = √(153.44 N^2 + 1033.03 N^2)
Fnet = √1186.47 N^2
Fnet ≈ 34.44 N
To find the direction of the net force, we use the inverse tangent (arctan) function.
Direction of the net force (θ) = arctan(Net Fy / Net Fx)
θ = arctan(32.16 N / -12.39 N)
θ ≈ -69.16°
Therefore, the magnitude of the net force is approximately 34.44 N and its direction is roughly 69.16° clockwise from the x-axis.