A lever is 5m long. The distance from fulcrum to the weight to be lifted is 1m. If the worker pushes on the opposite end with 400N, what is the max weight that can be lifted?
l=5m
h=1m
F=400N
F_g=?
torque=F_g*l
I first must find torque so that I can plug it into the above equation.
F=torque/r
400N=torque/5m
=2000Nm
thus
200Nm =F_g(5m)
F_g=400N
is this correct? Or would r=1m? Does the 1m in the question matter for this problem? Or is it just extra information?
No. The lever distance is from the point of force to the fulcrum. Here, it is 4 m (not five meters as you did).
400*4=torque= W*1
W= 1600N
If you want to make it more simple !!, here's another answer.
By newtons first law, the sum of all static forces at a balanced system is zero. Otherwise it will be moving...!!!
Add all vertical forces:
Here you have to consider a hidden force R form Reaction at support in opposite direction of forces (here at fulcrum acting upwards)
W + 400 - R = 0
R = W+400
Add all Torques about any point within or outside the lever. Better choose the point of unknown force or fulcrum to make it simple, as in the previous answer. Distances are based on the point you select.
Let's take about point of unknown force W:
W*0 + 400*5 - R*1 = 0 ;
R = 400*5
plug in value of R from previous step:
(W+400) = 400*5 ;
W = 400*5 - 400 = 2000-400 = 1600 N
quite simple OK. Now try with any other point.
Well, well, well! Look who's trying to lift some weights! Let's break it down, shall we?
First things first, the distance from the fulcrum to the weight is indeed relevant. It determines the lever arm, which affects the torque. In this case, you have a 5m lever with a load positioned 1m away from the fulcrum.
Now, to calculate the maximum weight that can be lifted, we need to consider the equilibrium of torques. The worker applies a force of 400N on the opposite end of the lever. To maintain equilibrium, the torque created by the worker's force must equal the torque created by the weight.
The torque equation is Torque = Force x Lever Arm. In this case, the worker's torque would be 400N x 5m = 2000Nm, as you correctly calculated.
Here's where it gets interesting. The torque created by the weight depends on its unknown force, let's call it F_g (short for Funny_gravity). The weight's torque would be F_g x 1m, based on its distance from the fulcrum. As torque must be balanced, we can set up an equation:
2000Nm = F_g x 1m
Solving for F_g, we find that F_g = 2000N.
So, the maximum weight that can be lifted in this scenario is a whopping 2000 Newtons! Keep flexing those physics muscles, my friend!
In this situation, the 1m distance mentioned in the question does matter. It is the distance from the fulcrum to the weight to be lifted, which is known as the "load arm" or "resistance arm". The other end of the lever where the worker pushes is the "effort arm".
To solve this problem, you correctly used the concept of torque. Torque is the product of the force and the perpendicular distance from the point of rotation (fulcrum) to the line of action of the force. In this case, the torque is given by the formula: torque = force x distance.
Given:
Length of lever (l) = 5m
Length of load arm (h) = 1m
Force applied by the worker (F) = 400N
First, let's calculate the torque applied by the worker. The force applied by the worker (F) multiplied by the length of the effort arm (5m) gives us the torque:
Torque = F x l = 400N x 5m = 2000Nm
Now, we can use the torque to calculate the maximum weight that can be lifted. The weight being lifted (F_g) is located on the load arm, so we can use the formula:
Torque = F_g x h
Plugging in the values, we have:
2000Nm = F_g x 1m
To solve for F_g, we divide both sides of the equation by 1m:
F_g = 2000Nm / 1m = 2000N
So, the maximum weight that can be lifted with a lever length of 5m, an effort force of 400N, and a load arm length of 1m is 2000N.