Forklift stopping distance and tipping condition analysis

[nineeyes]

I attached the problem and diagram.

This problem has been bothering me a bit. A friend of mine told me I could solve for the stopping distance by just looking at the crate.
I got
x: $$\mu*N_w = M*a$$
y: $$N_w-W_w=0$$
and from this got
$$a=-9.66\frac{ft}{s^2}$$

I used the acceleration and given inital velocity of 10 ft/s and final velocity of 0 ft/s and solved for the stopping distance through

$$s=\frac{-v_0^2}{2*a}$$

Assuming I approached this correctly, I wasn't sure how to check if it doesn't tip. I thought that I could just take the moment about a point and solve for $$N_B$$, I did

$$\sum M_A=N_B*7-W_G*4+W_W*3-a*3=0$$
and got

$$N_B=210.15$$ lbs. $$\therefore$$ it does not tip

Is that right? I also was wondering if I have to consider frictional force in the moment equation.
Thanks in advance.

 

[member 553083]

Yes, your approach is correct. You do not need to consider frictional force in the moment equation since the crate is not moving. You simply need to make sure that when you calculate the normal force of the base (N_B) it is greater than the weight of the crate (W_G). If N_B > W_G then the crate will not tip.

 

[thepatientmental]

Your approach to solving the stopping distance is correct. Using the given initial and final velocities, you can solve for the acceleration and then use that to calculate the stopping distance. However, to accurately check if the forklift will tip, you need to consider the moments of all the forces acting on it.

In your calculation for the moment about point A, you did not include the moment due to the frictional force. This force can contribute to the tipping moment and should be included in your calculation. You can calculate the frictional force using the coefficient of friction and the normal force N_B.

Once you have calculated the frictional force, you can then recalculate the moment about point A and see if it is still balanced. If it is not balanced, then the forklift will tip. If it is balanced, then the forklift will not tip.

In addition, it is important to note that the tipping point is not just dependent on the weight of the crate and the forklift, but also on the weight distribution of the load. If the load is not evenly distributed, it can create an unbalanced moment and cause the forklift to tip.

Overall, your approach to solving the problem is correct, but it is important to consider all the forces and moments in order to accurately determine if the forklift will tip or not.