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?
1 answer:
Answer:
(a) 104 N
(b) 52 N
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
You might be interested in
Answer:
Electric potential energy at the negative terminal:
Explanation:
When a particle with charge travels across a potential difference
, then its change in electric potential energy is
In this problem, we know that:
The particle is an electron, so its charge is
We also know that the positive terminal is at potential
While the negative terminal is at potential
Therefore, the potential difference (final minus initial) is
So, the change in potential energy of the electron is
This means that the electron when it is at the negative terminal has of energy more than when it is at the positive terminal.
Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is