A box is pulled up a rough ramp that makes an angle of 22 degrees with the horizontal surface. The surface of the ramp is the x-plane.There is a free body diagram drawing on a block on a 22 degree incline with 5 force vectors. The first force is pointing up and parallel to the surface of the incline, labeled F Subscript T Baseline. The second force is pointing down and parallel to the surface of the incline, labeled F Subscript f Baseline = 14.8 N. The third force is pointing away from and perpendicular to the surface of the incline, labeled F Subscript N Baseline = 65 N. The fourth force is pointing into and perpendicular to the surface of the incline, labeled F Subscript g y Baseline. The fifth force is pointing straight down to the center of the earth, labeled F Subscript g Baseline = 70 N. There is another force parallel to the surface of the incline and down the hill starting at the tip of the F Subscript g y Baseline vector; it is labeled F Subscript g x Baseline.


What is the magnitude of the force of tension if the net force in the x direction of 98 N?


57 N

139 N

178 N

183 N

PLEASE EXPLAIN HOW YOU GOT ANSWER!

A box is pulled up a rough ramp that makes an angle of 22 degrees with the horizontal surface The surface of the ramp is the xplaneThere is a free body diagram class=

Respuesta :

Magnitude of the force  of tension: 139 N

Explanation:

The surface of the ramp here is assumed to be the positive x-direction.

To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.

There are three forces acting along the x-direction:

  • The force of tension, [tex]F_T[/tex], acting up along the plane
  • The force of friction, [tex]F_f=14.8 N[/tex], acting down along the plane
  • The component of the weight in the x-direction, [tex]F_{gx}[/tex], acting down along the plane

We know that the magnitude of the weight is

[tex]F_g=70.0 N[/tex]

So its x-component is

[tex]F_{gx}=F_g sin \theta =(70.0)(sin 22^{\circ})=26.2 N[/tex]

The net force along the x-direction can be written as

[tex]F_x = F_T-F_f-F_{gx}[/tex]

And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:

[tex]F_T=F_x+F_f+F_{gx}=98+14.8+26.2=139 N[/tex]

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Answer:

139N

on edg

Explanation: