Question

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?

Answer 1

As we know that
Fx = Fcos30.0deg 
60.0N = Fcos30.0deg 
F = 60.0N/cos30.0deg 
F = 69.3 N
 Fy = Fsin30.0deg 
Fy = 69.3 N sin 30.0 deg 
Fy = 34.6 N
hope it helps

Answer 2

Answer:

a.69.3 N

b.34.7 N

Step-by-step explanation:

We are given that

Angle of inclination of the ramp=$20^{\circ}$

Force F makes an angle with ramp=$30^{\circ}$

The component  of F parallel to the ramp =$60 N$

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

We know that

$F_x=F cos 30^{\circ}$

$F=\frac{F_x}{cos 30^{\circ}}$

$F=\frac{60}{\frac{\sqrt3}{2}}$

$F=40\sqrt3=69.3N$

b.We have to find $F_y$ perpendicular to the ramp.

$F_y=Fsin 30^{\circ}$

$F_y=69.3\times \frac{1}{2}=34.7 N$