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 60N? b) How large will the component F, then be?

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lublana lublana · Answer 1 · 2019-10-29T09:47:52+01:00

Answer:

a.69.3 N

b.34.7 N

Step-by-step explanation:

We are given that

Angle of inclination of the ramp=[tex]20^{\circ}[/tex]

Force F makes an angle with ramp=[tex]30^{\circ}[/tex]

The component of F parallel to the ramp =[tex]60 N[/tex]

a.We have to find the value of when its horizontal component is 60 N.

We know that

[tex]F_x=F cos 30^{\circ}[/tex]

[tex]F=\frac{F_x}{cos 30^{\circ}}[/tex]

[tex]F=\frac{60}{\frac{\sqrt3}{2}}[/tex]

[tex]F=40\sqrt3=69.3N[/tex]

b.We have to find [tex]F_y[/tex] perpendicular to the ramp.

[tex]F_y=Fsin 30^{\circ}[/tex]

[tex]F_y=69.3\times \frac{1}{2}=34.7 N[/tex]