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princejan7
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Homework Statement
http://postimg.org/image/7vpxry28t/
Can someone explain how they calculated the force R representing the distributed load?
Did they even make use of the value "800N/m" from the question?
Yes, they used the 800, but they obtained the force R result a stupid (IMHO) way. The way I would have done it would have been to note that the average distributed load over the length of the member is 600 N/m. If we multiply that by the length of the member (0.6), we get 360 N. They did something like the following: the minimum distributed load over the length of the member is 400 N/m, so this contributes 400 (0.6) = 240N. Over and above that, the remainder of the load varies from 0 at the left end to 400 at the right end (400 to 800, minus the 400 already accounted for). The average of this excess is (0+400)/2 = 200. The load contribution of this excess is (400/2)(0.6)=120N. The total load R is again 360 N. As I said, their method is kinda stupid.princejan7 said:Homework Statement
http://postimg.org/image/7vpxry28t/
Can someone explain how they calculated the force R representing the distributed load?
Did they even make use of the value "800N/m" from the question?
The formula for calculating force with distributed load is Force = Load per unit length x Length.
A distributed load is a force that is spread out over a certain area or length rather than being concentrated at a single point.
Load per unit length can be determined by dividing the total load by the total length.
Sure, let's say you have a 10 ft beam with a distributed load of 1000 lbs/ft. The total load would be 10,000 lbs (1000 lbs/ft x 10 ft). If you want to find the force at a specific point, say 5 ft from one end, the force would be 5,000 lbs (1000 lbs/ft x 5 ft).
Yes, it is important to also consider the direction and distribution of the load, as well as any support or constraints on the beam. These factors can affect the calculation and should be taken into account for accurate results.