Find the expression of force required to slide down the body at inclined plane?
Answers
f we neglect friction between the body and the plane - the force required to move the body up an inclined plane can be calculated as
Fp = W h / l
= W sin α
= m ag sin α (1)
where
Fp = pulling force (N, lbf)
W = m ag
= gravity force - or weight of body (N, lbf)
h = elevation (m, ft)
l = length (m, ft)
α = elevation angle (degrees)
m = mass of body (kg, slugs)
ag = acceleration of gravity (9.81 m/s2, 32.174 ft/s2)
By adding friction - (1) can be modified to
Fp = W (sin α + μ cos α)
= m ag (sin α + μ cos α) (2)
where
μ = friction coefficient
Example - Pulling Force on Inclined Plane
A body with mass 1000 kg is located on a 10 degrees inclined plane. The pulling force without friction can be calculated as
Fp = (1000 kg) (9.81 m/s2) sin(10o)
= 1703 N
= 1.7 kN
Answer:
fn = normal force on the body by the surface of incline
m = mass of the body
θ = angle of incline
μ = coefficient of kinetic friction
a = acceleration of the body
fk= kinetic frictional force
perpendicular to incline surface , force equation can be given as
fn= mg Cosθ eq-1
kinetic frictional force is given as
fk= μ fn
using eq-1
fk = μ mg Cosθ eq-2
parallel to incline , force equation is given as
mg Sinθ - = ma
using eq-2
mg Sinθ - μ mg Cosθ = ma
a = g (Sinθ - μ Cosθ )
hope it may helpful
