contestada

celera
An
A force F = 89.2 N acts at an angle a = 38°
with respect to the horizontal on a block of
mass m = 26.8 kg, which is at rest on a
horizontal plane.
The acceleration of gravity is 9.81 m/s2.
If the static frictional coefficient is pls
0.74, what is the force of static friction?
=

celera An A force F 892 N acts at an angle a 38 with respect to the horizontal on a block of mass m 268 kg which is at rest on a horizontal plane The accelerati class=

Respuesta :

leena

Hi there!

We know that the MAXIMUM static friction force to be overcome is:

Fs (Force of Static Friction) = μN (Normal Force)

The normal force is equal to the weight of the block AND the vertical component of the applied force, so:

N = Mg + Fasinθ

Plug in the given values:

N = 26.8(9.81) + (89.2)sin(38) = 317.825 N

Multiply by the force of static friction to derive the static force:

0.74 * 317.825 = 325.191 N <-- MAXIMUM STATIC FRICTION FORCE

The force applied, if not greater than this value, will result in an EQUAL and OPPOSITE static frictional force. Thus:

Fs = Horizontal component of the applied force

Fs = 89.2cos(38) = 70.291 N

Otras preguntas