mdcstudent Posts: 1, Reputation: 1 New Member #1 Feb 15, 2005, 07:36 PM
Physics (Current and Parallel Forces)
Hi HelpDesk:

Please I need help how to figure out these problems:

1) An auto engine of mass 295Kg is located 1.0 m from one end of a 4.0
M work-bench. If the uniform bench has a mass of 45.0Kg, what weight
Must each end of the bench support?

2) There is a graphic like:

_____|_________________|_____
10.m |A 200m |B 1.0m

In the graphic shows 4.0m long, weighs 1550N and is made of uniform material. A weight of 245N hangs 1.00m from the end. Find the tension of each support cable.

Any solutions or feed backs it will be appreciated it

Thanks

James D.
[email protected]
 kokikiko Posts: 2, Reputation: 1 New Member #2 Sep 20, 2009, 10:46 PM

my ans for number one problem:
F1=243.75
F2=96.25
 kokikiko Posts: 2, Reputation: 1 New Member #3 Sep 20, 2009, 10:51 PM

here's my solution for problem number 1
F1(4m)=295kg(3m)+45(2m)
F1=295(3)+45(2) divided by 4

F1=243.75

for F2:

F1+F2=F3+F4
F2=295+45-243.75
F2=96.25
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #4 Sep 21, 2009, 08:30 AM

Well... what I got for 1) is:

Since the mass of the bench is 45 kg, it exerts a force of 450 N (taking acceleration due to gravity 10 m/s^2)

So, both ends will firstly support 450/2 = 225 N

Ok, now the engine.

Take the pivot to be at the left end of the bench and find the moments. Let's assume the engine is 1 m from the left end.

Clockwise moment = Anticlockwise moment
(2950 x 1) = (F x 4)
F = 2950/4 = 737.5 N

So, total weight on the right end support = 737.5 + 225 = 962.5 N

Now, weight of engine on the left is given by 2950-737.5 = 2212.5 N

Total weight on left support = 2212.5+225 = 2437.5 N

Weights: 962.5 N and 2437.5 N

Now try the second one, using the same principle.

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