Calculating support forces on each tire

AI Thread Summary
To determine the support forces at each wheel of a stationary truck with two crates, the total mass of the truck and crates must be calculated first. The gravitational force acting on the system is then found by multiplying the total mass by gravity. Since the load is not uniformly distributed, the center of gravity needs to be identified to accurately calculate torque. The sum of all forces must equal the total weight for equilibrium, which will help in finding the support forces on each tire. This approach ensures a correct assessment of the forces acting on the truck's wheels.
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Homework Statement



Two uniform crates of 350kg are placed in the bed of a stationary 1500kg truck. Determine the support forces at each of the wheels

The dimensions are only around the horizontal axis

Homework Equations


Fx = n1 + n2 -M1g -m2g-m3g

T= Dmg


The Attempt at a Solution



Thats about where I am at, I am just looking for a hint,
 
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Add mass of crates and truck -> find gravitational force -> divide by 4 (force is split over 4 tires)
 
tramar said:
Add mass of crates and truck -> find gravitational force -> divide by 4 (force is split over 4 tires)
\

If it was uniform that would be right but its not uniform. This is what I came up with:

Find center of gravity, Calculate torque, using that the sum of all forces has to equal total mass it should cancel out, any thoughts?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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