A block weighing #6 kg# is on a plane with an incline of #(3pi)/8# and friction coefficient of #2/5#. How much force, if any, is necessary to keep the block from sliding down?

2 Answers
Apr 10, 2017

The force is #=63.3N#

Explanation:

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Let the force be #F#

The frictional force is

#mu_s=F_r/N#

#N=mgcostheta#

#F_r=mu_s*mg costheta#

Resolving in the direction parallel to the plane #↗^+#

The force is

#F=F_r+ mgsintheta#

#F=mu_s mgcostheta+mgsintheta#

#=2/5*6*9.8cos(3/8pi)+6*9.8*sin(3/8pi)#

#=(9+54.3)N#

#=63.3N#

The force is #=63.3N#

Apr 10, 2017

#"it is necessary a force 45.38 Newtons"#

Explanation:

#"given data :"#
#m:6 " "kg" mass of block"#
#g=9.81 " "N/(kg)#
#beta=(3pi)/8#
#mu_k=2/5#

#"The block appears as if it were given on an inclined plane."#
#"Please notice that the weight of of block is perpendicular"#
#"to the ground."#

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#"You should split the weight of block into two components "#

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#"The green vector "G_x " is parallel to the inclined plane."#

#"The orange vector "G_y " is perpendicular to the inclined plane."#

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#G_x=mg*sin beta#

#G_x=6*9.81*sin beta#

#color(green)(G_x=54.38 )" "N#

#G_y=mg*cos beta#

#G_y=6*9.81* cos beta#

#color(orange)(G_y=22.52)" " N#

#"since it is perpendicular to the surface ,The " G_y " component causes friction."#

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#color(red)(F_f=mu_k*G_y)#

#F_f=2/5*22.52#

#F_f=9" "N#

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#"Since "G_x " greater than "F_f ," block is sliding down."#
#F_("net")=G_x-F_f=54.38-9#

#F_("net")=45.38 " "N#

#"it is necessary a 45 Newtons of force at opposite direction"#
#"so that it does not slide."#