Is the lever formula ##F*a=Q*b## or ##Q*a=F*b## is it the same thing ? thanks
I think that nobody has responded, because nobody is quite sure what you are talking about. Can you give us some indication of what F, Q, a, and b are supposed to be?
a-arm of force
b-arm of burden
V a b V
If this is the situation, where 'F' is the applied force, 'a' is the distance between the point where F is applied and the pivot point (fulcrum), 'Q' is the weight of the load, and 'b' is the distance between the load and the fulcrum, then in order for the force applied to exactly balance the load, the equation is:
F*a = Q*b
If you think about it, this is clearly NOT the same as F*b = Q*a, which would be wrong. It's just basic algebra that these two equations are not equivalent. If you have 3*2 = 6*1, this is clearly not the same thing as saying that 3*1 = 6*2. The second equation is clearly wrong.
However, if you are looking for a physical explanation for why the lever formula is true, then it has to do the concept of torque. When the situation is in equilibrium (i.e. the lever is stable and is not falling in either direction), the sum of the torques (τ) around the pivot point is zero. Now, the magnitude of the torque due to the applied force F is given by the magnitude of F multiplied by the perpendicular distance between the pivot point and the point where F is applied. In this case, this distance is a. Hence, the magnitude of the torque due to force F is τF = F*a. Similarly, the torque due to the load is equal to the load force multiplied by the perpendicular distance between the load and the pivot point. τQ = Q*b.
When the system is in equilibrium, the net torque is zero, which means that the two torques are equal in magnitude and opposite in direction, hence they cancel each other out. If they're equal in magnitude, then τF = τQ, and so F*a = Q*b.
If you haven't encountered the concept of torque and you don't know what I'm talking about, then don't worry about it. Just ignore the previous two paragraphs.
there is an arm of length 200 mm which is pivoted at 40 mm. that means we are getting one lengths along pivot as 160 and 40=200. suppose i am applying a force of 10N on extreme end of shorter length. what force i will get on the other extreme end. you can see example at following image
Welcome to PF,
Generally, if you have a new topic, it's better to start a new thread rather than add on to somebody else's.
In any case, just use the formula that I discussed above. You have both lengths, and one out of the two forces. There is not reason why you can't use that equation to solve for the second force.
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