A box is being pushed to the left parallel to the x-axis by a force of 23.0 N. A rope provides a second force of 33.8 N on the box, directed to the right at an angle of 39.0 deg. from the x-axis. What is the magnitude of the net force acting on the box ?
-F1+F2•cos α=-23 +33.8•cos39°= 3.27 N
That is not the answer. That's what I thought it was but its not.
F(x)= -F1+F2•cos α=
=-23 +33.8•cos39°= 3.27 N
F(y) = F2•sin α =33.8•sin39°= 21.27 N
F(net) =sqrt{F(x)²+F(y)²}= sqrt{3.27²+21.27²}=21.52 N
To find the magnitude of the net force acting on the box, we can use vector addition.
The force of 23.0 N directed to the left along the x-axis can be represented as F1 = -23.0 N, where the negative sign indicates its direction.
The force of 33.8 N provided by the rope is directed to the right at an angle of 39.0 degrees from the x-axis. To find the x-component of this force, we can use trigonometry. The x-component (horizontal component) of this force is given by F2x = F2 * cos(theta).
F2 = 33.8 N (magnitude of the force)
theta = 39.0 degrees (angle from the x-axis)
F2x = 33.8 N * cos(39.0 degrees)
Now, we can add the x-components of the forces to find the net force along the x-axis:
Fnetx = F1 + F2x
Finally, to find the magnitude of the net force, we can use the Pythagorean theorem:
|Fnet| = sqrt(Fnetx^2 + F2y^2)
Let's calculate the values step by step:
- For F2x:
F2x = 33.8 N * cos(39.0 degrees)
F2x ≈ 25.91 N (rounded to two decimal places)
- For Fnetx:
Fnetx = -23.0 N + 25.91 N
Fnetx ≈ 2.91 N (rounded to two decimal places)
- For |Fnet|:
|Fnet| = sqrt(2.91 N^2 + F2y^2)
Note that we still need to calculate the y-component of the force provided by the rope. The y-component (vertical component) of this force is given by F2y = F2 * sin(theta) where
theta = 39.0 degrees (angle from the x-axis)
F2y = 33.8 N * sin(39.0 degrees)
F2y ≈ 21.03 N (rounded to two decimal places)
Now, we can substitute this value in the equation to find |Fnet|:
|Fnet| = sqrt(2.91 N^2 + (21.03 N)^2)
|Fnet| ≈ 21.49 N (rounded to two decimal places)
Therefore, the magnitude of the net force acting on the box is approximately 21.49 N.