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[cm304eex]

Hi guys had this question asked in our lecture today, I have no idea where to start as our teacher explained it at about 100mph. so any help would be great!! headbang

A bar 1.5m long is pivoted at its centre. A downward force of 90N is aplied at right angles 0.2m from the end.

Calculate the downward force at to be applied at right angles to the bar at the opposite end to prevent it from rotating. neglect the weight of the bar.

helpheadbang

 

[SPECIAL LOCATION]

Hi guys had this question asked in our lecture today, I have no idea where to start as our teacher explained it at about 100mph. so any help would be great!! headbang A bar 1.5m long is pivoted at its centre. A downward force of 90N is aplied at right angles 0.2m from the end.

Calculate the downward force at to be applied at right angles to the bar at the opposite end to prevent it from rotating. neglect the weight of the bar.

helpheadbang

Bar 1.5 pivoted at the centre so there is 0.75m either side of pivot...

90M is applied 0.2m from one end...

e.g. (0.75-0.2) = 0.55 from the pivot...

Remember equal force equal length = Balance

Half the force but double the length on one side will also balance

Force & length are proportional to each other either side of pivot!

So 90N is 0.55 from pivot..

How much force is needed 0.75 on other side of pivot?

90N x 0.55m MUST equal (x?)N x 0.75m

so

90x.55= 49.5

49.5 = (x?)N x 0.75m

49.5/0.75 = 66N

so 66N at one end will balance 90N 0.2m from other end!



is that OK?

its nearly 1.00 am & I have had a few Guiness DrinkGuiness DrinkGuiness Drink

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