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Question Number 89300 by I want to learn more last updated on 16/Apr/20

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Commented by I want to learn more last updated on 16/Apr/20

Determine the minimum horizontal force required to maintain a 10kg box on a ramp inclined at 30° to the horizontal if the coefficient of friction is 0.2

Commented by mr W last updated on 17/Apr/20

Commented by mr W last updated on 17/Apr/20

N=mg cos α+F sin α  f=mg sin α−F cos α  f=μN  mg sin α−F cos α=μ(mg cos α+F sin α)  mg(sin α−μ cos α)=F(μsin α+cos α)  mg(tan α−μ)=F(μ tan α+1)  ⇒F=((mg(tan α−μ))/(μ tan α+1))=((10×10(tan 30°−0.2))/(0.2×tan 30°+1))  =33.83 N

N=mgcosα+Fsinαf=mgsinαFcosαf=μNmgsinαFcosα=μ(mgcosα+Fsinα)mg(sinαμcosα)=F(μsinα+cosα)mg(tanαμ)=F(μtanα+1)F=mg(tanαμ)μtanα+1=10×10(tan30°0.2)0.2×tan30°+1=33.83N

Commented by I want to learn more last updated on 17/Apr/20

Thanks sir. I appreciate.

Thankssir.Iappreciate.

Answered by 242242864 last updated on 17/Apr/20

Horizontal force = F  Resolving along the plane  Fcos ∝+uR=mgsin ∝.......(i)  Vertically the plane,  Fsin ∝+mgcos ∝=R  ⇒Fcos ∝+u(Fsin ∝+mgcos ∝)=mgsin ∝  but ∝=30^°   Fcos 30^° +0.2(Fsin 30^° +10×9.8cos 30°)=10×9.8sin 30^°   0.87F+0.1F+16.97=49  F= ((32.03)/(0.97))  F= 33.02N

Horizontalforce=FResolvingalongtheplaneFcos+uR=mgsin.......(i)Verticallytheplane,Fsin+mgcos∝=RFcos+u(Fsin+mgcos)=mgsinbut∝=30°Fcos30°+0.2(Fsin30°+10×9.8cos30°)=10×9.8sin30°0.87F+0.1F+16.97=49F=32.030.97F=33.02N

Commented by I want to learn more last updated on 17/Apr/20

Thanks sir, i appreciate

Thankssir,iappreciate

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