Friction caused by acceleration on a rocket sled

Question

I'm trying to quantify some intuition I have about friction and rocket sled.

Suppose a rectangular sled with length $l$, weight $W$ on a surface with friction coefficient $mu$. A rocket attached height $h$ from the ground on the middle of the sled exerts force $H$ causing the sled to accelerate, call the right pad Pad1 and left pad Pad2. Find the difference of friction between Pad1 and Pad2.
(assume the weight of the sled is big enough to not go flying, flipping or something unusual, if will only accelerate to the right)
Of course, the sum of friction on Pad1 and Pad2 would be (weight of the sled) x (coefficient of friction). And if H was not there, the difference would be zero because the weight of the sled will be balanced evenly on the two pads ie) same normal force. But since there is that $H$ force, I think I have to account for that, so I tried using my intuition.
My intuition is telling me Pad1 should have greater friction, I think $H$ should push down on Pad1 and it will push up on Pad2, with the same magnitude, causing the balance of friction to "tip over" to Pad1, but the sum of the pad would stay the same as $Wmu$. I can't quite quantify how much of this "unbalance" occurs, though. any help?

Bob D · Accepted Answer

Your intuition is correct.
If the sled is not to tip over or otherwise lose contact with the ground, the sum of the vertical forces on the sled must equal zero and the sum of the moments about any point on the sled must be zero.
Neglecting the weight of the rocket, summing the vertical forces give us
$$-W+R_{1}+R_{2}=0$$
$$R_{2}=W-R_{1}$$
Where $R_{1}$ and $R_{2}$ are the upward vertical reactions at Pads 1 and 2 respectively.
Again, neglecting the weight of the rocket, the sum of the moments about contact surface of Pad 2 with the surface is (clockwise moments being positive):
$$Hh-R_{1}L+frac{WL}{2}=0$$
$$R_{1}=frac{W}{2}+frac{Hh}{L}$$
Plugging this into the vertical forces equations to determine $R_2$
$$R_{2}=frac{W}{2}-frac{Hh}{L}$$
So the reaction force on Pad 1 is greater than that on Pad 2. Since the reaction forces are the forces normal to the friction surface, the friction force on Pad 1 is greater than Pad 2.
So your intuition is correct. The friction force is greater on Pad 1 then Pad 2.
Hope this helps.

R.W. Bird · Answer

You are concerned about rotational stability. You have the vertical force equation: $N_f$ + $N_r$ - mg = 0 with two unknown N's at the front and rear. Combine that with a torque equation taken about the front support: Hh + $N_r$L – mg(L/2) = 0. You need $N_r$ to be greater than zero.

Answered by R.W. Bird on May 14, 2021

Friction caused by acceleration on a rocket sled

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