clear
Hello everyone,
I have this problem which driving me crazy
The problem is Number 21. I already solved Number 20.

20.) You're using a wedge to split a log. You are hitting the wedge with a large hammer to drive it into the log. It takes a force of 2000 N to push the wedge into the wood. If the wedge moves 0.2 into the log, how much work have you done on the wedge?

this is the answer We know the work we do on an object is calculated by multiplying the applied force and the distance it travels in the direction of motion. In this case, all the force is along the direction of motion, so the work is (2000N)*(0.2 m) = 400 J

21) the wedge in problem 20 acts like, slowly splitting the wood apart as it enters the log. the work you do in the wedge pushing it into the log, is the work it does on the wood, separating its two halves. if the two halves of the wedge only separate by distance of 0.05 m while the wedge travels 0.2m into the log, how much force is the wedge exerting on the two halves of the log to rip them apart.

Can you guys tell me how to calculate it ?
Thank you all
Robittybob1
QUOTE (clear+Mar 1 2012, 01:36 AM)
Hello everyone,
I have this problem which driving me crazy
The problem is Number 21. I already solved Number 20.

20.) You're using a wedge to split a log. You are hitting the wedge with a large hammer to drive it into the log. It takes a force of 2000 N to push the wedge into the wood. If the wedge moves 0.2 into the log, how much work have you done on the wedge?

this is the answer We know the work we do on an object is calculated by multiplying the applied force and the distance it travels in the direction of motion. In this case, all the force is along the direction of motion, so the work is (2000N)*(0.2 m) = 400 J

21) the wedge in problem 20 acts like, slowly splitting the wood apart as it enters the log. the work you do in the wedge pushing it into the log, is the work it does on the wood, separating its two halves. if the two halves of the wedge only separate by distance of 0.05 m while the wedge travels 0.2m into the log, how much force is the wedge exerting on the two halves of the log to rip them apart.

Can you guys tell me how to calculate it ?
Thank you all

Work in = work out
Pressure times distance1 = X (Pressure) * Distance2

400/Distance 2 = X (Pressure)
But this pressure is split 50:50 I think.
clear
Thank you for responding.
As I know the final answer is 8000 N.
However I don't know how to calculate it.
If you don't mind telling me how?

Thanks again
Robittybob1
QUOTE (clear+Mar 1 2012, 03:41 AM)
Thank you for responding.
As I know the final answer is 8000 N.
However I don't know how to calculate it.
If you don't mind telling me how?

Thanks again

It is the slope of the wedge the ratio of the distances moved reflects the shape. 4:1 slope.
freenergy
x= .05/.2 (2000n)

right?
Robittybob1
QUOTE (freenergy+Mar 1 2012, 11:26 AM)
x= .05/.2 (2000n)

right?

NO. Does that when worked out come to 8000?
Cwilm
FxD= work, the work you found in the first is the same work done in the second, however, the distance is ow .05m. Therefore your new formula is..

F x .05= 400j

Divide 400 by .05 and you'll get 8000N
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