A box on a 20 degree incline is shown with vectors radiating from a point in the center of the box. The first vector points up and parallel to the surface of the incline, labeled F Subscript f s Baseline. A second vector points toward the center of the earth, labeled F Subscript g Baseline = 735 N. A third vector is perpendicular to and away from the surface of the incline from the point, labeled F Subscript N Baseline. A fourth vector is broken into 2 components, one parallel to the surface and down the incline, labeled F Subscript g x Baseline, and one perpendicular to the surface and into the surface, labeled F Subscript g y Baseline.

A box at rest on a ramp is in equilibrium, as shown.

What is the force of static friction acting on the box? Round your answer to the nearest whole number. N

What is the normal force acting on the box? Round your answer to the nearest whole number.

Respuesta :

The images to solve this problem is in the attachment.

Answer: [tex]F_{fs}[/tex] = 671.0 N; [tex]F_{N}[/tex] = 300 N

Step-by-step explanation: From the image in the attachment and knowing that the box is in equilibrium, i.e., the "sum" of all the forces is 0, it is possible to conclude that:

[tex]F_{fs}[/tex]  = [tex]F_{gx}[/tex] and [tex]F_{N}[/tex] = [tex]F_{gy}[/tex]

Using trigonometry, shown in the second attachment, the values for each force are:

  • Force of Static Friction

sin 20° = [tex]\frac{F_{gx} }{F_{g} }[/tex]

[tex]F_{gx}[/tex] = [tex]F_{g}[/tex]. sin(20)

[tex]F_{gx}[/tex] = 735.0.913

[tex]F_{gx}[/tex] = 671.0

  • Normal Force

cos 20° = [tex]\frac{F_{gy} }{F_{g} }[/tex]

[tex]F_{gy}[/tex] = [tex]F_{g}[/tex]. cos (20)

[tex]F_{gy}[/tex] = 735.0.408

[tex]F_{gy}[/tex] = 300

The force of static friction is 671N and normal force is 300N

Ver imagen cristoshiwa
Ver imagen cristoshiwa

Answer:

Static force is 251 and the Normal force is 691.

Step-by-step explanation:

Hope this helps!! Have a great day!!    :)