Question

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.

Answer 1

Static force is 251 and the Normal force is 691.

Explanation:

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

Answer 2

The images to solve this problem is in the attachment.

Answer:
`F_(fs)` = 671.0 N;
`F_(N)` = 300 N

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:

`F_(fs)` =
`F_(gx)` and
`F_(N)` =
`F_(gy)`

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

• Force of Static Friction

sin 20° =
`(F_(gx) )/(F_(g) )`

`F_(gx)` =
`F_(g)`. sin(20)

`F_(gx)` = 735.0.913

`F_(gx)` = 671.0

• Normal Force

cos 20° =
`(F_(gy) )/(F_(g) )`

`F_(gy)` =
`F_(g)`. cos (20)

`F_(gy)` = 735.0.408

`F_(gy)` = 300

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