Two forces act on an object. The first force has a magnitude of 19.0 N and is oriented 58.0° counter-clockwise from the x axis, while the 2nd force has x- and y-components (-25.0 N, 16.5 N). Express the magnitude and direction of the net force.

Question

Two forces act on an object. The first force has a magnitude of 19.0 N and is oriented 58.0° counter-clockwise from the x axis, while the 2nd force has x- and y-components (-25.0 N, 16.5 N). Express the magnitude and direction of the net force.

Answer 1

F1 + F2 = 19[58o] + -25+16.5i =

19*Cos58+i19*sin58 -25 + 16.5i =
10.1 + 16.1i - 25 + 16.5i=-14.9+32.6i,Q2
= 35.8N.[65.4o] N. of W. = 35.8N[144.2o]
CCW.

Answer 2

To find the magnitude and direction of the net force acting on the object, we need to calculate the vector sum of the two forces.

First, let's convert the second force's components (-25.0 N, 16.5 N) into magnitude and direction form.

The magnitude of the second force can be found using the Pythagorean theorem:

Magnitude = √((-25.0 N)^2 + (16.5 N)^2)
Magnitude = √(625 N^2 + 272.25 N^2)
Magnitude = √(897.25 N^2)
Magnitude ≈ 29.96 N

We can use trigonometry to find the direction of the second force. The direction angle θ can be found using the arctangent function:

θ = arctan(16.5 N / -25.0 N)
θ ≈ -32.01°

Now, let's add the two forces together to find the net force.

First, we need to resolve the first force into x and y components. The x-component can be found using the cosine function, and the y-component can be found using the sine function:

F1x = 19.0 N * cos(58.0°)
F1x ≈ 9.88 N

F1y = 19.0 N * sin(58.0°)
F1y ≈ 15.96 N

Now, let's sum the x and y components of the two forces:

Fnetx = F1x + F2x
Fnetx ≈ 9.88 N + (-25.0 N)
Fnetx ≈ -15.12 N

Fnety = F1y + F2y
Fnety ≈ 15.96 N + 16.5 N
Fnety ≈ 32.46 N

The magnitude of the net force can be found using the Pythagorean theorem:

Magnitude = √(Fnetx^2 + Fnety^2)
Magnitude = √((-15.12 N)^2 + (32.46 N)^2)
Magnitude = √(228.12 N^2 + 1054.4916 N^2)
Magnitude = √(1282.6116 N^2)
Magnitude ≈ 35.79 N

The direction of the net force can be found using the inverse tangent function:

θ = arctan(Fnety / Fnetx)
θ = arctan(32.46 N / -15.12 N)
θ ≈ -64.82°

So, the magnitude of the net force acting on the object is approximately 35.79 N, and the direction is approximately 64.82° counter-clockwise from the x-axis.