A lift of mass $520kg$ is supported by a steel rod attached to its bottom. The rod can withstand the maximum thrust $15,400N$. The lift can accelerate at $2.5ms^{-2}$ and decelerate at $7.8ms^{-2}$. Find the maximum allowed load in the lift.
This question is confusing me as I am not very comfortable with the normal reaction force and any help would be appreciated. I have attached my working, but it isn’t correct, and I am unsure where I have gone wrong.
Many thanks.
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Let "M" be the maximum load, in kilograms. Carrying that load, the combined mass of elevator and load is M+ 520 kg. Since "force equals mass times acceleration", F= ma, at 2.5 m/s^2 acceleration, the force is F= (M+ 520)(2.5)= 2.5M+ 1300 N. We are told that "The rod can withstand the maximum thrust 15,400N" so to find the maximum load we must solve 2.5M+ 1300= 15400. This is done entirely in terms of mass and acceleration so I see no need to multiply by "g" to get weight.
Let the maximum load be $M$ kg, so the total mass (lift + load) is $520+M$. The greatest force delivered by the rod when the lift is accelerating at maximum rate is when the lift is accelerating upwards. Then we require $$15400-(520+M)g\ge(520+M)2.5$$ Similarly, for deceleration, the worst case is when the lift is decelerating at 7.8 with the rod having to overcome gravity (ie slowing down as it is coming down). We need $$15400-(520+M)g\ge(520+M)7.8\quad(*)$$ So $(*)$ is the more restrictive condition and gives $$M(g+7.8)\le15400-520g-4056$$ Taking $g=9.81$ that gives $$M=\frac{6242.8}{17.61}=354.5\text{ kg}$$ Of course, in practice one would allow a considerable safety margin, so the allowed load would be smaller.