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، کاز 37o = 0.8، g = 10 ms ثانیې-2، μk = 0.2)
پیژندل شوی:
ډله (m) = 2 kg
سرعت د جاذبې له امله (g) = ۱۰ متره/ ثانیه2
Block’s د وزن (w) = m g = (2)(10) = 20 Newton
ګناه ۳۷o = 0.6
کاس 37o = 0.8
Coefficient of the حرکي رګونه (µk) = 0.2،XNUMX
The y-component of the weight (wy) = د کاس ایکس اینوم ایکسo = (۱۹.۶)(۰.۵) = ۹.۸ نیوټن
The x-component of the weight (wx) = w sin θ = (20)(sin 37) = (20)(0.6) = 12 Newton
the normal force (N) = wy = ۱۰ نیوټن
پراري : The external force (F)
د حل :
wx = ۱۰ نیوټن
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 :
ف + فk - wx = 0
ف = wx - fk
F = 12 – 1.6
F = 10.4 نیوټن
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.
پیژندل شوی:
The coefficient of the static friction (µs) = 0.4،XNUMX
زاویه (θ) = 45o
د جاذبې قوې له امله سرعت (g) = 10 m/s2
Block’s mass (m) = 2 kilogram
Block’s weight (w) = m g = (2 kg)(10 m/s2) = ۱۰ کیلوګرامه متر مکعب/ثانیه2 = ۱۰ نیوټن
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
پراري : The magnitude of the force F
حل:
Block starts to slide up, if F ≥ wx + fs.
The x-component of the weight :
wx = ۹.۸√۳ نیوټن
the y-component of the weight :
wy = ۹.۸√۳ نیوټن
نورمال ځواک :
ن = wy = ۹.۸√۳ نیوټن
The force of the static friction :
fs = μs ن = (۰.۵)(۴۰√۲) = ۲۰√۲
The magnitude of the force F so that the block starts to slide up :
F ≥ wx + fs
F ≥ ۴۰√۲ + 4√2
F ≥ 14√2 نیوټن
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