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September 27th, 2016, 12:18 PM   #1
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moments & forces with spring balance

Hello All

see the image of the caravan .if it is attached to a car, the caravan according to the safety regulations should exert a force of 690 N (6,9x 100 N) exerting downward on the car's tow hitch.
Before the person fastens the caravan, he checks the magnitude of the force by holding the caravan in equilibrium with a spring balance. the spring balance shows a force of 690 N. See image. also see the pivot point S and the center of gravity Z of the caravan The the caravan is drawn to scale.

a What is the distance from the pivot point S of the two forces acting on the caravan.

b Determine from the figure, the mass of the empty caravan.
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September 28th, 2016, 12:30 AM   #2
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My son(14) got this question in his test paper & he completely left it.
I just want to know how to go about on this so that I can back-explain it to my son.
Thanks for any insights.
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September 28th, 2016, 02:00 AM   #3
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The key point here is "The caravan is drawn to scale". Therefore, the answer to a) is obtained by measuring the distance between S and the tow hitch with a ruler. If there is an additional scale added to the diagram, then you will need to use the scale to convert the measured result into the correct scale. For example, if the diagram is 1:100 scale and you measure 10cm, then the real result is 1000cm = 10m.

For part b) You need to understand that in the case where the caravan is attached to a car, the moment caused by the force at the tow hitch is going to balance the moment caused by the weight of the caravan acting vertically downwards through the caravan's centre of mass (point Z). Because S is a pivot point, that is a dead giveaway that we are actually interested in the moments.

According to the law of moments, a system is in rotational equilibrium if the sum of the clockwise moments balance with the sum of the anticlockwise moments.

Let's call the distance measured in part a) 'A'. The moment for the tow-hitch is 690N $\displaystyle \times$ A.

Let's call the weight of the caravan W. The moment caused by this weight is W $\displaystyle \times$ B, where B is the measured distance between S and the point vertically below Z on the tow hitch.

Equating the moments, we get:

$\displaystyle W \times B = 690 \times A$

It looks to me like B is about one quarter of the distance of A (check it with a ruler!). Therefore:

$\displaystyle W = 690 \times 4 = 2760$N

The weight of the caravan is equal to the mass multiplied by the gravitational acceleration constant for Earth:

$\displaystyle W = M \times g$


$\displaystyle M = \frac{W}{g} = \frac{2760}{9.81} = 281.346$ kg

You will need a calculator for that Sometimes on non-calculator exam papers the exam paper will state that g should be approximated from 9.81 $\displaystyle m/s^2$ to 10 $\displaystyle m/s^2$. If that is the case, the answer is very easy because you just divide the result by 10 instead:

$\displaystyle M = \frac{W}{g} = \frac{2760}{10} = 276$ kg
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September 28th, 2016, 10:19 AM   #4
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Lots of Thanks Benit13
it would help me help him. !
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