A man pulls his child in a sled at a constant velocity. The child and sled have a combined mass of 60 kg. The force applied on the handle is 200N at an angle of 15. Draw a free body diagram. What is the normal force acting on the sled?

With this kind of problem, it's very important that you keep track of your sign conventions. I set the y direction perpendicular (outwards from) to the ramp as positive and x direction parallel down the ramp as negative. Thus, #vecg < 0# in this convention.

Here's the free body diagram I came up with (treat the sled and child as one unified object):

Remember that #vecF_(g,||) = mvecgsintheta# is the parallel force component due to gravity that opposes the pulling force and #vecF_(g,_|_) = mvecgcostheta# is the perpendicular force component due to gravity that opposes the normal force.

Therefore, numerically, #vecF_(pu ll)# and #vecF_k# oppose each other, and numerically, #vecF_N# and #mvecgcostheta# oppose each other, and we can write a sum of the forces:

#sum_i vecF_(x,i) = -vecF_k + vecF_(pu ll) + mgsintheta = 0#

#sum_i vecF_(y,i) = vecF_N + mgcostheta = 0#

where #vecF_k# is the reacting force of the kinetic friction, #vecF_(pu ll) = "200 N"# is the force used to pull the sled, and #vecF_(g,||) = mgsintheta# is the force component of gravity parallel to the ramp. Note that #veca_x = 0# due to the constant velocity.

The plus and minus signs in the equations are such that the forces numerically oppose each other when they add later on. I know that #vecF_N > 0# and #mu_k > 0#, so #vecF_k > 0# (the minus sign accounts for left being negative). The reason why #mgsintheta# in the same direction has a plus sign is because #vecg < 0# already.

You can see we can solve for the normal force right away.

#color(blue)(vecF_N) = -mgcostheta#

#= -("60 kg" xx -"9.81 m/s"^2 xx cos(15^@)) = color(blue)("568.54 N")#

Once we have that, we should recall the definition of the kinetic friction force:

#vecF_k = mu_kvecF_N#,

where #mu_k# is the coefficient of kinetic friction.

Thus, the coefficient of kinetic friction is acquired like so:

#-vecF_k + vecF_(pu ll) + mgsintheta = 0#

#-vecF_k = -vecF_(pu ll) - mgsintheta = -mu_kvecF_N#

#color(blue)(mu_k) = (vecF_(pu ll) + mgsintheta)/vecF_N#

#= (("200 N") + ("60 kg")(-"9.81 m/s"^2)sin(15^@))/("568.54 N")#

#= color(blue)(0.0838)#

And the force of kinetic friction would be:

#color(blue)(vecF_k) = vecF_(pu ll) + mgsintheta#

#= "200 N" + ("60 kg")(-"9.81 m/s"^2)(sin15^@)#

#=# #color(blue)("47.66 N")#