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?
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.
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
To find the force that the man must exert on the rope to keep the sled moving up the ramp at a constant velocity, we need to analyze the forces acting on the sled.
First, let's break down the gravitational force into its components. The force of gravity on the sled is given by:
F_gravity = m * g
where m is the mass of the sled (80 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The force of gravity can be broken down into two components: one parallel to the incline and one perpendicular to the incline. The force component parallel to the incline, F_parallel, is given by:
F_parallel = m * g * sin(theta)
where theta is the angle of the incline (25 degrees). Substituting the values, we have:
F_parallel = 80 kg * 9.8 m/s^2 * sin(25 degrees)
Next, let's consider the frictional force acting on the sled. The frictional force, F_friction, is given by:
F_friction = coefficient of friction * normal force
The normal force, F_normal, is the force perpendicular to the incline and is equal in magnitude to the weight component perpendicular to the incline. The weight component perpendicular to the incline, F_perpendicular, is given by:
F_perpendicular = m * g * cos(theta)
Substituting the values, we have:
F_perpendicular = 80 kg * 9.8 m/s^2 * cos(25 degrees)
Finally, we can use F_friction and F_parallel to find the net force acting on the sled along the incline. Since the sled is moving at a constant velocity, the net force is zero. Therefore, we have:
Net force = F_parallel - F_friction
Setting the net force to zero, we can solve for F_parallel:
F_parallel = F_friction
Substituting the values, we have:
80 kg * 9.8 m/s^2 * sin(25 degrees) = 0.70 * 80 kg * 9.8 m/s^2 * cos(25 degrees)
Now, you can calculate the value of F_parallel, which is the force that the man must exert on the rope to keep the sled moving up the ramp at a constant velocity.