
There are many kinds of machines that we use everyday to do work or help us accomplish a task. All machines are fundamentally built up of combinations of six categories of simple machines: lever, pulley, wheel and axle, ramp, wedge, and screw. The mechanical advantage of a machine indicates how it changes the input force into an output force. The lever is a simple machine that multiplies force by the ratio of the input arm to the output arm.

machine, simple machine, mechanical advantage, input force, output force, lever, fulcrum, input arm, output arm


$$MA=\frac{{F}_{o}}{{F}_{i}}$$


$$M{A}_{lever}=\frac{{L}_{i}}{{L}_{o}}$$



Review problems and questions 

 Is a car a machine? Is it a simple machine? Why or why not?

A car is definitely a machine, because it can change the direction and/or magnitude of force (by converting forces from the pistons into force turning the wheels) and it can do work (by changing the kinetic energy of the car). It is not a simple machine, because it cannot be described as only one of the six different types of simple machines. As we will learn later, it is a combination of simple machines called a compound machine.

 How are first, second, and thirdclass levers different?

The difference among the three classes of levers comes from the location of the fulcrum relative to the input and output forces (or the effort and the load). In a firstclass lever, the fulcrum is located between the input and output forces. In a secondclass lever, the output force is located between the input force and the fulcrum. In a thirdclass lever, the input force is located between the output force and the fulcrum.

 The MegaRontastic Machine, resembling a fancy broom, is advertised on latenight television as a device to help you do household chores with less effort. Maximillian bought one and tested its mechanical advantage to see whether it actually worked as advertised. What measured values of mechanical advantage would justify the product claims about the MegaRontastic Machine?

The product claim is that you expend less effort to do the same output work or produce the same output force. That means that your input force is less than the output force. Since the mechanical advantage is given by MA = F_{o}/F_{i}, the product claims would be justified only if MA > 1.

 Prince Ludwig II creates his own version of a machine that will lift a 25 kg bucket of water. Igor tries out the device and determines that, to lift the water by 2 m, he must pull the input rope by 10 m while applying 50 N of force. What is the mechanical advantage of Prince Ludwig II’s machine? (Assume that no energy is lost to friction.)

Answer: MA = 4.1 Asked: mechanical advantage MA of the machine Given: output mass m_{o} = 25 kg; input force F_{i} = 60 N Relationships: MA = F_{o}/F_{i}; F_{w} = mg Solve: The output force is the weight of the bucket:$${F}_{o}={m}_{o}g=(25\text{kg})(9.8{\text{m/s}}^{2})=245\text{N}$$The mechanical advantage is the ratio of the output force to the input force:$$MA=\frac{{F}_{o}}{{F}_{i}}=\frac{245\text{N}}{60\text{N}}=4.1$$Note that we did not need the information on either the input or output distance!

 Astrid has two slotted masses and a meter ruler. She wants to hang the 100 g from one end of the meter ruler (at the 0 cm mark), hang the 300 g mass from the other end (at the 100 cm mark), and place the meter ruler on a knife edge to balance the lever. At what mark on the ruler should she place the knife edge? The mass of the ruler itself is negligible.

Answer: The knife edge (fulcrum) should be placed at the 75 cm mark of the meter ruler. Asked: the mark on the ruler that corresponds to the input length of the lever, L_{i} Given: length of the ruler, d = 100 cm; input mass m_{i} = 100 g; output mass m_{o} = 300 g Relationships: If the lever is in balance, then the torques on each side of the fulcrum must be equal. Torque is force times distance, τ = FL. Weight is F = mg. Solve: Take the 100 g mass to be the “input” arm of the lever and the 300 g mass to be the output arm. The sum of the input and output arms must equal 100 cm:$${L}_{i}+{L}_{o}=100\text{cm}$$The input torque must be balanced by the output torque for the lever to be in equilibrium:$${m}_{i}g{L}_{i}={m}_{o}g{L}_{o}$$Divide both sides by m_{o}g and cancel terms to solve for L_{o}:$$\begin{array}{ccc}\frac{{m}_{i}\overline{)g}{L}_{i}}{{m}_{o}\overline{)g}}=\frac{{\overline{)m}}_{o}\overline{)g}{L}_{o}}{{\overline{)m}}_{o}\overline{)g}}& \Rightarrow & {L}_{o}=\frac{{m}_{i}}{{m}_{o}}{L}_{i}\end{array}$$Substitute this expression for L_{o} into the first expression and solve for L_{i}:$$\begin{array}{ccc}{L}_{i}+\frac{{m}_{i}}{{m}_{o}}{L}_{i}={L}_{i}\left(1+\frac{{m}_{i}}{{m}_{o}}\right)=100\text{cm}& \Rightarrow & {L}_{i}=\frac{100\text{cm}}{1+({m}_{i}/{m}_{o})}\end{array}=\frac{100\text{cm}}{1+(100\text{g}/300\text{g})}=75\text{cm}$$


Zing is a safecracker who wants to lift a 150 kg safe full of loot, but he can only exert 490 N of force. He has a 60cmlong steel rod as well as a rock he can use as a fulcrum.
 What minimum mechanical advantage does he need?
 How far from the safe should the rock be placed to provide this minimum mechanical advantage?
 What torque does Zing apply to lift the safe?
 What is the net torque about the fulcrum?

Answer:  Zing needs a minimum MA of 3.0.
 The rock (the fulcrum) should be placed 15 cm from the safe.
 Zing applies 221 N m of torque.
 The net torque is zero.
Solution:  Asked: mechanical advantage MA of the lever
Givens: applied force F_{i}; mass of safe, m = 150 kg Relationships: MA = F_{o}/F_{i} Solve:$$MA=\frac{{F}_{o}}{{F}_{i}}=\frac{mg}{{F}_{i}}=\frac{\left(150\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)}{490\text{N}}=3.0$$Answer: Zing needs a minimum MA of 3.0.  Asked: the distance from the load to the fulcrum, L_{o}
Givens: MA = 3.0; length of lever, (L_{o} + L_{i}) = 60 cm Relationship: MA = L_{i}/L_{o} Solve:$$\begin{array}{l}{L}_{o}+{L}_{i}=60\text{cm}\Rightarrow \text{}{L}_{i}=\left(60\text{cm}\right){L}_{o}\\ MA=\frac{{L}_{i}}{{L}_{o}}=\frac{\left(60\text{cm}\right){L}_{o}}{{L}_{o}}=3.0\\ \left(60\text{cm}\right){L}_{o}=3.0\text{}{L}_{o}\\ 60\text{cm=4.0}{L}_{o}\\ {L}_{o}=\frac{60\text{cm}}{4.0}=15\text{cm}\end{array}$$Answer: The rock (the fulcrum) should be placed 15 cm from the safe.  Asked: The input torque applied by Zing, τ_{i}
Givens: F_{i} = 490 N; length of lever, (L_{o} + L_{i}) = 60 cm; L_{o} = 15 cm Relationship: The input torque is given by τ_{i} = +L_{i}F_{i}. Solve:$$\begin{array}{l}{L}_{i}=60\text{cm}{L}_{o}=60\text{cm}15\text{cm=45cm}\\ {\tau}_{i}={L}_{i}{F}_{i}=\left(0.45\text{m}\right)\left(490\text{N}\right)=221\text{Nm}\end{array}$$Answer: Zing applies 221 N m of torque.  Asked: the net torque torque τ_{net}
Givens: the input torque τ_{i}; L_{o} = 15 cm; mass of safe, m = 150 kg Relationship: $${\tau}_{net}={\tau}_{i}+{\tau}_{o}=+{L}_{i}{F}_{i}{L}_{o}{F}_{o}=0$$Solve:$${\tau}_{net}={\tau}_{i}{\tau}_{o}=\left(221\text{Nm}\right)\left(0.15\text{m}\right)\left(150\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)=0\text{Nm}$$Answer: The net torque is zero.

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