Essential Physics

Section 1 review

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. Read the text aloud

Read the text aloud

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

M A = \frac{F_{o}}{F_{i}}

M A_{l e v e r} = \frac{L_{i}}{L_{o}}

Review problems and questions

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

How are first-, second-, and third-class levers different?

The Mega-Rontastic Machine, resembling a fancy broom, is advertised on late-night 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 Mega-Rontastic Machine?

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 kg) (9.8 {m/s}^{2}) = 245 N

The mechanical advantage is the ratio of the output force to the input force:

M A = \frac{F_{o}}{F_{i}} = \frac{245 N}{60 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 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_og and cancel terms to solve for L_o:

\begin{matrix} \frac{m_{i} g L_{i}}{m_{o} g} = \frac{m_{o} g L_{o}}{m_{o} g} & \Rightarrow & L_{o} = \frac{m_{i}}{m_{o}} L_{i} \end{matrix}

Substitute this expression for L_o into the first expression and solve for L_i:

\begin{matrix} L_{i} + \frac{m_{i}}{m_{o}} L_{i} = L_{i} (1 + \frac{m_{i}}{m_{o}}) = 100 cm & \Rightarrow & L_{i} = \frac{100 cm}{1 + (m_{i} / m_{o})} \end{matrix} = \frac{100 cm}{1 + (100 g / 300 g)} = 75 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 60-cm-long steel rod as well as a rock he can use as a fulcrum.
1. What minimum mechanical advantage does he need?
2. How far from the safe should the rock be placed to provide this minimum mechanical advantage?
3. What torque does Zing apply to lift the safe?
4. 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: $M A = \frac{F_{o}}{F_{i}} = \frac{m g}{F_{i}} = \frac{(150 kg) (9.8 {m/s}^{2})}{490 N} = 3.0$ 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: $\begin{array}{l} L_{o} + L_{i} = 60 cm \Rightarrow L_{i} = (60 cm) - L_{o} \\ M A = \frac{L_{i}}{L_{o}} = \frac{(60 cm) - L_{o}}{L_{o}} = 3.0 \\ (60 cm) - L_{o} = 3.0 L_{o} \\ 60 cm = 4.0 L_{o} \\ L_{o} = \frac{60 cm}{4.0} = 15 cm \end{array}$ 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_iF_i.
Solve: $\begin{array}{l} L_{i} = 60 cm - L_{o} = 60 cm - 15 cm = 45 cm \\ τ_{i} = L_{i} F_{i} = (0.45 m) (490 N) = 221 N m \end{array}$ 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: $τ_{n e t} = τ_{i} + τ_{o} = + L_{i} F_{i} - L_{o} F_{o} = 0$ Solve: $τ_{n e t} = τ_{i} - τ_{o} = (221 N m) - (0.15 m) (150 kg) (9.8 {m/s}^{2}) = 0 N m$ Answer: The net torque is zero.


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