# Force and resultant

1. Nov 16, 2009

### MorallyObtuse

Is a.) correct?
Great help is required in the other two.

1. The problem statement, all variables and given/known data

a.) Fine salt is deposited from a negligible height at a rate of 12kgs-1 onto a conveyor belt moving at speed 8.0ms-1.Calculate the force required to keep the belt moving at this velocity.

b.) A small aircraft of mass 1200 kg which is flying horizontally at 150 ms-1 is taken into a vertical circular path of radius 500m and it is kept in this path until the aircraft makes a complete loop. If the pilot has a mass of 72kg, calculate the force exerted on him by the seat when the aircraft is top and bottom of the loop.
Clueless, much assistance needed here, please.
c.) A lorry of total mass 12000 kg enters a horizontal circular bend of radius 350 m at a constant speed of 45kmh-1.Calculate the magnitude and the direction of the resultant horizontal force acting on the lorry as it travels around the bend
b. Calculate the angle to which the road must be banked in order to provide the same resultant without the need to rely on friction.
Clueless, much assistance needed here, please.

2. Relevant equations

a.) F α rate of change of momentum

3. The attempt at a solution

a.) F α m/t(v - u)
F α 12(8)
F α 96 N

2. Nov 16, 2009

### cepheid

Staff Emeritus
There are some problems with your solution to part a, although I think you probably got the right answer. First of all, I think (but I am not sure) that you have written that the force is proportional to the rate of change of momentum, using the Greek letter alpha as a symbol for "proportional to." There are three problems with this:

1. The symbol for proportional to is not the same as the symbol for alpha (even though they look quite similar when handwritten):

$$\propto \, \, \, \, \, \textrm{(proportional to)}$$

$$\alpha \, \, \, \, \, \textrm{(alpha)}$$ ​

2. Alpha doesn't show up well in the default font used on the forums, and it just ends up looking like the letter 'a' (which makes your post confusing).

3. F is not merely proportional to the rate of change of momentum, it is in fact exactly equal to it (Newton's 2nd Law). I.e. the constant of proportionality is just 1. So you should really write:

$$\textbf{F} = \frac{d\textbf{p}}{dt}$$ ​

As for your solution, if the rate of change of momentum is constant, then we can write

$$F = \frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t}v$$ ​

where the mass flow rate $\Delta m / \Delta t$ and the velocity v are both given in the problem. This is what your equations should say. I'm not sure what v and u mean in the equations you wrote, but it seems that you somehow plugged in the right numbers.

3. Nov 16, 2009

### MorallyObtuse

Yep, I meant proportional to. And thanks.