The above is the solution,But I dont understand please explain with details dont

Question

The above is the solution,But I dont understand please explain with details dont just give me solution

Solve first two equations for N and Fn' to obtain Fn? = m [4t2cos10degree/30 - gsin10degree] N = m [4t2sin10degree/30 + gcos10degree] Condition for slipping: Square both sides and solve for t: t = 5.58s The flatbed truck starts from rest on a road whose constant radius of curvature is 30 m and whose bank angle is 10degree. If the constant forward acceleration of the truck is 2 m/s2, determine the time t after the start of motion at which the crate on the bed begins to slide. The coefficient of static friction between the crate and truck bed is mu s = 0.3, and the truck motion occurs in a horizontal plane. Ans. t = 5.58 s

Explanation / Answer

the fbd is quite explained,N perpendicular to truck base,mg acting verticaly downward,n dir..towards centre of curvature,first eqn is equillibrium along y axis,taking compnents of all the forces along y axis.

2nd eqn...sum of forces along x axis eqyated to m*aN,aN is centripetal accn=V^2/R,V=2*t,v=u+at,t =time when slipping starts,equations solved to get Fn',Ft.now for slipping to start frictional force=uS*N which is dir opposite to Fn' should equate Ft +Fn',the case for impending friction

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