# Force problem

• Sep 12th 2008, 10:38 AM
windseaker
Force problem
If you have 117cu.ft of dirt@ 120lb/cuft moving on a conveyor belt (which is 22" wide) at 10" per min, what force (and what particular force) would be hitting the wall at the end of the conveyor?
and if it was hitting an object attached at the wall which is 12"high,22" wide, and just say 14" deep(to wall)?
• Sep 12th 2008, 10:47 AM
topsquark
Quote:

Originally Posted by windseaker
If you have 117cu.ft of dirt@ 120lb/cuft moving on a conveyor belt (which is 22" wide) at 10" per min, what force (and what particular force) would be hitting the wall at the end of the conveyor?
and if it was hitting an object attached at the wall which is 12"high,22" wide, and just say 14" deep(to wall)?

Hint: Use the Impulse-Momentum theorem: $\bar{F} \Delta t = m \Delta v$
where $\bar{F}$ is the average force you are looking for. You can easily compute the $\Delta v$ in ft/s. Now imagine how much mass is hitting the wall in 1 second of time. This will give you your average force. And notice that we don't care about how high the dirt is for the first part, but that we need to know it for the second part. So if the dirt is higher than the 12" object then only that fraction of the dirt that is up to 12" high will impact the object on the wall.

-Dan
• Sep 12th 2008, 11:15 AM
windseaker
Force Problem
OK, 10"/mim convert to ft/s
10"/12=.833 then /by 60sec will =.0138 ft/sec

now 117cuft(120lb./cuft)= 14,076lb.

finally 14,076lb.(.0138ft/sec)=194.3 lbft/sec??
• Sep 12th 2008, 11:25 AM
topsquark
Quote:

Originally Posted by windseaker
OK, 10"/mim convert to ft/s
10"/12=.833 then /by 60sec will =.0138 ft/sec

now 117cuft(120lb./cuft)= 14,076lb.

finally 14,076lb.(.0138ft/sec)=194.3 lbft/sec??

Okay, you bring up an interesting point. Your answer is correct if all of the material strikes the wall in exactly 1 second. But there is no way to specify what the height of the dirt is. So instead of all the dirt impacting the wall in 1 s the dirt might be more smeared out on the conveyor. If we don't know the height of the pile of dirt is then we can't figure out how much dirt is impacting on the wall in 1 s. (And note that the force is larger the higher the dirt gets because the impact takes less time.) I think the question is flawed for not having the height information.

-Dan
• Sep 12th 2008, 11:58 AM
windseaker
Force Problem
Alright , question b.

If the dirt is 117cuft (say at 22'wide,8' tall,8'long) moving on the conveyor toward an area of impact(say 22" wide and only12"high) and this area is elevated at 6" above the conveyor, what is the force impact at the given .0138 ft/sec?