A particle of weight 120N is placed on a plane inclined at 300 to the horizontal. If the plane has an efficiency of 60% to the floor what is the force required to push the weight uniformly up the plane?
50N
100N
175N
210N
Explanation
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It's just:
Eff=Workdone sin(teetah)/workdonex100/1
60=120x0.5/W x100
So,cross multiply
W=60x100/60
W=6000/60
W=100
Given Values
Weight (W) = 120 N
Angle of inclination (θ) = 30°
Efficiency (η) = 60% = 0.6
Calculate Force Required without Efficiency
Force required to push the weight up the plane without considering efficiency:
F = W sin(θ)
= 120 N x sin(30°)
= 120 N x 0.5
= 60 N
Calculate Force Required with Efficiency
Since the plane has an efficiency of 60%, the actual force required will be:
F_actual = F / η
= 60 N / 0.6
= 100 N
The force required to push the weight uniformly up the plane is 100 N.
Efficiency=60% or 0.6,angle of inclination=30°,weight load=120
vR=1/sin( θ)
vR=1/sin30°=1/0.5=2
ή=MA/VR
0.6=MA/2
MA=0.6×2
MA=1.2
MA=Load/effort
1.2=120/E
E=120/1.2
E=100N
