A man is dragging a trunk up the loading ramp of mover's truck. The ramp has a slope angle of 20 degree, and the man pulls upward with a force F whose direction makes an angle of 30 degree with ramp. a. How large a force F is a necessary or the component F, parallel to the ramp to be 80N? b. How large will the component F, then be?
To solve this problem, we can break down the force F into its components parallel and perpendicular to the ramp. Let's label the necessary component parallel to the ramp as F_parallel and the perpendicular component as F_perpendicular.
a. To find the necessary force component F_parallel, we can use trigonometry. The force F can be split into two components: F_parallel and F_perpendicular. The angle between F and the ramp is given as 30 degrees.
Using trigonometry, we can find F_parallel using the equation:
F_parallel = F * cos(theta)
where theta is the angle between F and the ramp.
In this case, F_parallel = 80 N and theta = 30 degrees.
Substituting the values into the equation, we get:
80 N = F * cos(30 degrees)
To solve for F, divide both sides of the equation by cos(30 degrees):
F = 80 N / cos(30 degrees)
Calculating that expression, we find that F is approximately 92.39 N.
Therefore, the necessary force component F_parallel to be 80 N is approximately 92.39 N.
b. To find the value of the perpendicular component F_perpendicular, we can use the relationship between F_parallel and F_perpendicular:
F_perpendicular = F * sin(theta)
In this case, theta is still 30 degrees and F is approximately 92.39 N.
Substituting the values into the equation, we get:
F_perpendicular = 92.39 N * sin(30 degrees)
Calculating that expression, we find that F_perpendicular is approximately 46.20 N.
Therefore, the perpendicular component F_perpendicular will be approximately 46.20 N.