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Grade 12Mechanics

A collar of mass m slides up the vertical shaft under the action of a force F of constant magnitude but variable direction. If θ = Kt, where K is a constant and if the collar starts from rest with θ = 0. Find the magnitude F of the force which will result in the collar coming to rest as θ reaches π/2. The coefficient of kinetic friction between collar and shaft is μ.

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Profile image of Aslan
8 Years agoGrade 12
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3 Answers

Profile image of Eshan
8 Years ago
Net force acting on the body upwards=Fcos\theta-mg-\mu (Fsin\theta)=ma

\implies \dfrac{dv}{dt}=\dfrac{Fcos\theta-mg-\mu (Fsin\theta)}{m}
=\dfrac{F}{m}(cos(Kt)-\mu sin(Kt))-g
\implies v=\int_0^{\pi/2}(\dfrac{F}{m}(cos(Kt)-\mu sin(Kt))-g)dt
\implies \int_0^0 dv=\int_0^{t_0}(\dfrac{F}{m}(cos(Kt)-\mu sin(Kt))-g)dt
\implies 0=\dfrac{F}{Km}(sin(Kt_0)-\mu +\mu cos(Kt_0))-gt_0

Since\dfrac{\pi}{2}=Kt_0\implies t_0=\dfrac{\pi}{2K}
\implies 0=\dfrac{F}{Km}(1-\mu )-\dfrac{g\pi}{2K}
\implies F=\dfrac{mg\pi}{2(1-\mu)}
Profile image of Gitanjali Rout
8 Years ago
JHVH\implies F=\dfrac{mg\pi}{2(1-\mu)}\implies 0=\dfrac{F}{Km}(1-\mu )-\dfrac{g\pi}{2K}\dfrac{\pi}{2}=Kt_0\implies t_0=\dfrac{\pi}{2K}Since\implies 0=\dfrac{F}{Km}(sin(Kt_0)-\mu +\mu cos(Kt_0))-gt_0\implies \int_0^0 dv=\int_0^{t_0}(\dfrac{F}{m}(cos(Kt)-\mu sin(Kt))-g)dt\implies v=\int_0^{\pi/2}(\dfrac{F}{m}(cos(Kt)-\mu sin(Kt))-g)dt=\dfrac{F}{m}(cos(Kt)-\mu sin(Kt))-g\implies \dfrac{dv}{dt}=\dfrac{Fcos\theta-mg-\mu (Fsin\theta)}{m}Fcos\theta-mg-\mu (Fsin\theta)=maNet force acting on the body upwards=
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Profile image of Forum Team
8 Years ago
 
 
Gitanjali, please stop copy pasting answers posted by other users. You will not be rewarded for these points and your account will be blocked due to this activity as it is against forum guidelines. Please check the guidelines here – Forum point Policy – https://www.askiitians.com/forum-point-policy/
 
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Forum Team