Question #f4617

I got #mu_s = 0.519#.

This is a classic free-body diagram you will need to know, an incline/ramp with a block acted upon.

It will contain a normal force, force due to gravity, sometimes force of friction, and in this case, also a pushing force upwards along the incline.

Based on this, we construct our sum of the forces for each coordinate direction.

For convenience, we rotate our coordinate system so that #x# is along the incline and #y# is perpendicular to the incline. Define down the ramp as positive and outwards from the ramp as positive (thus, the gravitational constant is #vecg > 0#).

The push was designed to stop the body (i.e. the block) from sliding, so the forces sum to #bb0#.

#sum_i vecF_(i,x) = vecF_g sintheta - vecF_p - vecF_s = 0#

#sum_i vecF_(i,y) = vecF_N - vecF_gcostheta = 0#

where:

• #vecF_g# is the force due to gravity, #mvecg#.
• #vecF_p# is the pushing force, #"10 N"#.
• #vecF_s# is the static friction force for sliding objects. We don't need to know this directly.
• #vecF_N# is the normal force, such that #vecF_s = mu_svecF_N#

Therefore:

#mvecgsintheta - "10 N" - mu_svecF_N = 0# #" "bb((1))#

#vecF_N - mvecgcostheta = 0# #" "bb((2))#

From this, we have enough information; we can use #(2)# to get #vecF_N# and plug into #(1)# to eliminate all unknowns.

#vecF_N = mvecgcostheta#

#=> mvecgsintheta - "10 N" - mu_s mvecgcostheta = 0#

Therefore, the coefficient of static friction is:

#=> color(blue)(mu_s) = (mvecgsintheta - "10 N")/(mvecgcostheta)#

#= (("3 kg")("9.81 m/s"^2)(sin45^@) - "10 N")/(("3 kg")("9.81 m/s"^2)(cos45^@))#

#= (("3 kg")("9.81 m/s"^2)(sqrt2/2) - "10 N")/(("3 kg")("9.81 m/s"^2)(sqrt2/2))#

#= color(blue)(0.519)#