Simple Force and Motion Problem - Explain the explanation please.

• tenbee
In summary, the conveyor belt of the incline will take longer to hit box B if box B remains stationary.
tenbee

Homework Statement

Half-Passage: In the packing industry, processing packages via conveyor belts is immensely practical and vital for operations. Maintaining packages in a neat and file line is no easy task, considering that many conveyor belts are non-linear and require the traversal of hills and valleys. As a general rule of thumb, boxes are kept no less than 5m apart so that collisions do not occur and skew the entire process. In the following diagram, box A is at the very top of the conveyor belt and box B is at the very bottom. The height of the conveyor belt system is h and the slops of the hill is r.

Question 2: Assume that the conveyor belt of the incline is distinct and moves separately from the conveyor belt on which B rests. If the conveyor belt underneath A breaks, how long will it take for box A to hit box B assuming that box B remains stationary and the coefficient of kinetic friction of the conveyor belt is 0.4? Also assume that the mass of box A is m. For simplicity, assume that the distance down the ramp is d.

The Attempt at a Solution

answer: {2d/[gsin(r) - 0.4gcos(r)]}1/2

Okay... I understand how to solve this equation, but I don't understand why it's '2d' instead of 'd'. A little help please : )

To find time...
a = v/t and v= d/t, so a = (d/t)/t
F = ma --> a = F/m
Ffr = µk*Fnorm

So for force --> (mgsinθ - mgcosθ)/m, then cancel the m --> a = (gsinθ - gcosθ)

(d/t)/t = (gsinθ - gcosθ) --> (d/t) = t(gsinθ - gcosθ) --> d = t*t(gsinθ - gcosθ --> d/(gsinθ - gcosθ) = t2 --> [d/(gsinθ - gcosθ)]1/2 = t

Where do they get 2d from?!?

Last edited:
Try setting a = 1/2 t^2, taken from x(t) = 1/2t^2 + vt + t_0

khemist said:
Try setting a = 1/2 t^2, taken from x(t) = 1/2t^2 + vt + t_0

Ahhh, I see - yes that works. Thank you!

When should I use x = x0 + v0t + 1/2at2 versus a = (d/t)/t?

x(t) is a position function. It implies constant acceleration and one can use it to determine the time it takes to get from A to B .

I would only use a = d/t^2 when you know the distance interval and time interval. It is really a = (d_2-d_1)/(t_2 - t_1)^2, which gives you an average acceleration.

Someone else might be able to answer that question a little better, I am not 100% on when to use it.

Actually the problem is wrong. If A is moving, the conveyor has a velocity v, and when it stops, it makes a get an disacelleration or an acelleration (only if static friction coefficient makes A don't slide in the conveyor, but kinetic does). This way A has a initial velocity v that increases or decreases by time, so the initial velocity is needed. But as the answer says, there is no v, so, assuming that v is TOO short and can be forgotten, we have:

We know the position/time function in the MUV:

$\Delta S = Vot - (1/2)at²$
$Vo=null$
$\Delta S = (1/2)at²$
$t=\sqrt{2 \Delta S/a}$

But a is the acceleration of gravity in the axe of the conveyor minus the disacxceleration of friction

$A = g.sin\alpha - A_{friction}$

$Fa = N.u = m.g.cos\alpha.u -> Fa = 0.4mgcos\alpha -> A_{ friction} = 0.4gcos\alpha$
$A = g.sin\alpha - 0.4gcos\alpha$
$t = \sqrt{ 2\Delta S/g.sin\alpha - 0.4gcos\alpha}$

Replace to get answer

What is force and motion?

Force and motion are related concepts in physics. Force is a push or pull applied to an object, while motion is the change in position or orientation of an object over time.

What is Newton's first law of motion?

Newton's first law of motion states that an object at rest will remain at rest and an object in motion will stay in motion with a constant velocity unless acted upon by an external force.

How do you calculate force?

Force can be calculated by multiplying an object's mass by its acceleration, as described by Newton's second law of motion: F = ma.

What are some examples of simple force and motion problems?

Some examples of simple force and motion problems include calculating the force required to accelerate an object, determining the distance traveled by an object given its initial velocity and acceleration, and finding the acceleration of an object based on the force applied to it.

How does force and motion apply to everyday life?

Force and motion are fundamental concepts that apply to many aspects of everyday life. For example, they can explain how a ball rolls down a hill, how a car moves when the gas pedal is pressed, and how a parachute slows down a skydiver's fall.

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