Question

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward with a force FS whose direction makes an angle of 30.0° with the ramp (Fig. E4.4). (a) How large a force FS is necessary for the component Fx parallel to the ramp to be 90.0 N?(b) How large will the component Fy perpendicular to the ramp be then?

Answer 1

To find the force FS necessary for the component Fx parallel to the ramp to be 90.0 N, we can use the equation Fx = FS * cos(30°) = 90.0 N. To find the component Fy perpendicular to the ramp, we can use the equation Fy = FS * sin(30°).

Step-by-step explanation:

To find the force FS necessary for the component Fx parallel to the ramp to be 90.0 N, we can use the equation Fx = FS * cos(30°) = 90.0 N. Rearranging the equation, FS = 90.0 N / cos(30°). Therefore, FS ≈ 103.9 N.

To find the component Fy perpendicular to the ramp, we can use the equation Fy = FS * sin(30°). Substituting the value of FS, Fy ≈ 103.9 N * sin(30°). Therefore, Fy ≈ 51.9 N.

Answer 2

(a) 104 N

(b) 52 N

Step-by-step explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.

The component of F perpendicular to the ramp is Fy.

(a)

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.

Resolve F into its x-component from Pythagorean theorem:

Fx=Fcos30°

Solve for F:

F= Fx/cos30°

Substitute for Fx from given data:

Fx=90 N/cos30°

=104 N

(b) Resolve r into its y-component from Pythagorean theorem:

Fy = Fsin 30°

Substitute for F from part (a):

Fy = (104 N) (sin 30°)

= 52 N