Regarding Friction/Normal force at 90(d) angle

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The discussion revolves around calculating the necessary acceleration of a truck for a person to remain stuck on its front after being hit. The forces acting on the person include gravity and the frictional force, represented by the coefficient of friction (μ). The user seeks clarification on how to determine the normal force (N) in this scenario, particularly when dealing with a 90-degree angle. The conversation also touches on the concept of pseudo forces in non-inertial frames, likening them to gravitational effects. Ultimately, the user expresses satisfaction with the explanation received.
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Homework Statement


I will try to explain it this way:
A truck hits a person, the person now is stuck on the trucks front.
Forces working on the person are Gravity and the force that the truck is applying.
The coefficient of the friction is \mu.

What should be the truck's acceleration in order for the person to stay "stuck" on its front.2. The attempt at a solution
\SigmaF\hat{y} = -mg+\mu*N
From what I understand we need \Sigma=0. I have never dealt with a problem in which the surface is placed at a 90 degrees angle. Could you guys please provide a solution/hint for what N is.

Thank you. :)
 
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N = m*a...
.
m = mass of the person
a = acceleration of the truck...
.
On a X-Y system...
If the truck is approaching along the x-axis from right to left...
Tilt your head 90 deg to left...
Follow the truck...
Now since you're in a non inertial frame...
The pseudo force acting is similar to what gravity would be if this all were friction calculated between an object and a leveled ground...
.
I'm curious... Are you planning to kill someone...
(O_o)
 
Well thanks, I got it now.

Ask my Classical Mechanics professor, I'm starting to suspect that he has killing tendencies.
 
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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