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Why is the output force always less than th input force in third clss lever?
Because the output distance is always greaterthan the input distance,and the product of (force) x (distance) is nearly the same on both ends.
Why is the output force always less than th input force in third class lever?
Because the output distance is always greaterthan the input distance,and the product of (force) x (distance) is nearly the same on both ends.
In a second class lever why must you exert input force over a greater distance?
If the input force is applied at a greater distance than the length of the effort arm is increased thereby reducing the effort.
How do you find the input force for a lever?
Ok, so a lever can be broken up into two 'sides' with a fulcrum in the middle. This idea simply utilizes the laws set forth for torque, or Force*distance. Static equilibrium (which would be when you input enough force on one side of the lever to balance the other) states the followingF1*D1 = F2*D2Starting from the left side of the lever, for have a force (F1) multiplied by the distance between that force and the fulcrum (D1). This can be set equal to the distance between the fulcrum and the second force, with this distance denoted as D2. If you want to know the input force, you need to know the other force, and both distances. Then you can simply divide. For example say want to know your input force, F2.F2 = (F1*D1)/D2Hope this helps
How do you calculate the input force you apply to bike pedals?
How you calculate the input force that you apply to bike pedals involves multiplying the force by the distance the object moves in the direction of the force. This is a part of the law of the lever.
Why is the output force always less than the input force in the third class lever?
Because the output distance is always greaterthan the input distance,and the product of (force) x (distance) is nearly the same on both ends.
Why is the output force always less than th input force in third clss lever?
Because the output distance is always greaterthan the input distance,and the product of (force) x (distance) is nearly the same on both ends.
Why is the output force always less than th input force in third class lever?
Because the output distance is always greaterthan the input distance,and the product of (force) x (distance) is nearly the same on both ends.
In a second class lever why must you exert input force over a greater distance?
If the input force is applied at a greater distance than the length of the effort arm is increased thereby reducing the effort.
How do you calculate a lever's mechanical advantage?
From the design of the lever (on paper), the mechanical advantage is effort arm/load arm which means Distance from pivot to the applied force/distance from pivot to the load The result of that is that the forces will have the reciprocal ratio, and the input force to the lever will be the output force/the Mechanical Advantage .
How do you find the input force for a lever?
Ok, so a lever can be broken up into two 'sides' with a fulcrum in the middle. This idea simply utilizes the laws set forth for torque, or Force*distance. Static equilibrium (which would be when you input enough force on one side of the lever to balance the other) states the followingF1*D1 = F2*D2Starting from the left side of the lever, for have a force (F1) multiplied by the distance between that force and the fulcrum (D1). This can be set equal to the distance between the fulcrum and the second force, with this distance denoted as D2. If you want to know the input force, you need to know the other force, and both distances. Then you can simply divide. For example say want to know your input force, F2.F2 = (F1*D1)/D2Hope this helps
What is input gear?
A Lever
What is input distance?
Input Distance is the distance the input force acts through.
Why is work input equal to the effort force times the effort distance?
That's the definition of "work" ... (force exerted) times (distance through which the force acts). If you push against the end of a lever with a force 'F' and move it through a distance 'D', then (F x D) is the work you put into the lever.
How do you calculate the input force you apply to bike pedals?
How you calculate the input force that you apply to bike pedals involves multiplying the force by the distance the object moves in the direction of the force. This is a part of the law of the lever.
How do you find the input force and output force?
let the input force be F1,and the distance between point of application of input force and the lever point is x1,similarly if output force iis F2,and distance of it's point of apllication is x2,then efficiency of the lever is (F2*x2)/(F1*x1) actually F*x gives the work done,and efficiency of any machine is output work/input work
Can a machine multiply input distance Input distance Input energy?
yes. yes. no
