1. A 2-kg block lies on a rough inclined plane at an angle 37o to the horizontal. Determine the magnitude of the external force exerted on the block, so the block is not slides down the plane. (sin 37o = 0.6, cos 37o = 0.8, g = 10 m.s-2, µk = 0.2)
Bekannt:
Mass (m) = 2 kg
Beschleunegung duerch Schwéierkraaft (g) = 10 m/s2
Block an Gewiicht (w) = mg = (2)(10) = 20 Newton
Ouni 37o = 0.6
Kosch 37o = 0.8
Coefficient of the kinetesch Reiwung (µk) = 0.2
The y-component of the weight (wy) = w fir 37o = (20)(0.8) = 16 Newton
The x-component of the weight (wx) = w sin θ = (20)(sin 37) = (20)(0.6) = 12 Newton
the normal force (N) = wy = 16 Newton
Gewënschte : The external force (F)
Léisung :
wx = 12 Newton
The force of the kinetic friction (fk) = µk N = (0.1)(16) = 1.6 Newton
The magnitude of the external force F exerted on the block :
F + fk - wx = 0
F = wx - fk
F = 12 – 1.6
F = 10.4 Newton
The external force F greater than 10.4 Newton.
2. Mass of a block = 2 kg, coefficient of static friction µs = 0.4 and θ = 45o. Determine the magnitude of the force F so the block start to slides up.
Bekannt:
The coefficient of the static friction (µs) = 0.4
Wénkel (θ) = 45o
Schwéierkraaftbeschleunigung (g) = 10 m/s2
Block’s mass (m) = 2 kilogram
Block’s weight (w) = m g = (2 kg)(10 m/s2) = 20 kg m/s2 = 20 Newton
The x-component of the weight (wx) = w sin θ = (20)(sin 45) = (20)(0.5√2) = 10√2 Newton
The y-component of the weight (wy) = w cos θ = (20)(cos 45) = (20)(0.5√2) = 10√2 Newton
Gewënschte : The magnitude of the force F
Léisung:
Block starts to slide up, if F ≥ wx + fs.
The x-component of the weight :
wx = 10√2 Newton
the y-component of the weight :
wy = 10√2 Newton
Déi normal Kraaft :
N = wy = 10√2 Newton
The force of the static friction :
fs = µs N = (0,4)(10√2) = 4√2
The magnitude of the force F so that the block starts to slide up :
F ≥ wx + fs
F ≥ 10√2 + 4√ 2
F ≥ 14√2 Newton
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