a man pulls an 80 kg sled up an incline that is at an angle of 25 degrees with the horizontal using a rope attached to the sled at an angle of 35 degrees. If the coefficient of friction between the sled and the incline is 0.70, what force must the man exert on the rope to keep the sled moving up the ramp at a constant velocity?

Answer 1

Ws = m*g = 80kg * 9.8N/kg = 784 N. = Wt.
of sled.

Fp = 784*sin25 = 331.3 N. = Force parallel to the incline.
Fn = 784*cos25 = 710.5 N. = Normal force = Force perpendicular to the incline.

Fk = u*Fn = 0.7 * 710.5 = 497.4 N. = Force of kinetic friction.

F-Fp-Fk = m*a
F-331.3-497.4 = m*0 = 0
F = 828.7 N. = Component of exerted force parallel to the incline.

Fex = F*cos25/cos35 =828.7*cos25/cos35=
917 N = Force exerted by the man.

Answer 2

α=25º, β =35º, m=80 kg, μ=0.7

Fcosβ - F(fr) - mgsin α = 0
N+Fsinβ – mg cos α = 0

N= mg cos α - Fsinβ
F(fr) = μN = μ(mg cos α – Fsinβ)

Fcosβ -μ(mg cos α – Fsinβ) - mgsin α = 0
Fcosβ -μ•mg cos α + μ Fsinβ - mgsin α =0

F(cosβ + μ•sinβ) =mg(sin α + μ•cos α)
F= mg(sin α + μ•cos α)/ (cosβ + μ•sinβ)=
=80•9.8(sin25+0.7cos35)/(cos25+0.7sin35) =
=64(0.42+0.573)/(0.906+0.401) =
=63.552/1.307=48.6 N