contestada

A 60kg block is at rest on a ramp inclined at 35 degrees.

A) Draw a FBD. Label normal, gravitational, and frictional force.
B) Determine the magnitude of the force of friction acting on the block
C) Determine the magnitude of the normal force acting on the block
D) Determine the coefficient of friction between the block and the ramp.

Respuesta :

Answer:

Explanation:

(a)Gravitational Force is W=mg

Frictional force is f

Normal force is N

(b)Since the block is in Equilibrium therefore from diagram we can see that  

[tex]f=mg\sin \theta [/tex]

[tex]f=60\times 9.8\times \sin 35[/tex]

[tex]f=337.26 N[/tex]

(c)Magnitude of Normal Force

[tex]N=mg\cos \theta =481.66 N[/tex]

(d)friction is also given by

[tex]f=\mu mg\cos \theta [/tex]

and [tex]f=mg\sin \theta [/tex]

thus [tex]tan \theta =\mu [/tex]

[tex]\mu =0.7[/tex]  

Ver imagen nuuk

Otras preguntas