Question

a 45.6-cm wheel is attached to an axle with a radius of 8.95 cm. when an effort force of 265 n turns the axle, the wheel moves a linear distance of 16.3 cm. the efficiency of the wheel and axle is 85.6 percent. a. what is the ima of the wheel and axle? b. what is the ma of the wheel and axle? c. what is the resistance force? d. how far did the axle turn when the wheel moved 16.3 cm?

Explanation:

Part a: IMA of Wheel and Axle

Step1: Recall IMA formula for wheel and axle

The Ideal Mechanical Advantage (IMA) of a wheel and axle is given by the ratio of the radius of the wheel (\(r_w\)) to the radius of the axle (\(r_a\)), i.e., \(IMA=\frac{r_w}{r_a}\).

Step2: Substitute known values

Given \(r_w = 45.6\space cm\) and \(r_a=8.95\space cm\), so \(IMA=\frac{45.6}{8.95}\approx5.095\) (rounded to three decimal places).

Step1: Recall efficiency formula

Efficiency (\(e\)) is related to Mechanical Advantage (MA) and IMA by the formula \(e=\frac{MA}{IMA}\times100\%\), so \(MA = e\times\frac{IMA}{100\%}\).

Step2: Substitute values

We know \(e = 85.6\%=0.856\) and from part (a) \(IMA\approx5.095\). So \(MA=0.856\times5.095\approx4.36\).

Step1: Recall MA formula for forces

Mechanical Advantage \(MA=\frac{F_r}{F_e}\), where \(F_r\) is resistance force and \(F_e\) is effort force. So \(F_r=MA\times F_e\).

Step2: Substitute values

We know \(MA\approx4.36\) (from part b) and \(F_e = 265\space N\). So \(F_r=4.36\times265 = 1155.4\space N\) (we can also use more precise \(MA\) value from part a: \(MA = 0.856\times\frac{45.6}{8.95}\), then \(F_r=0.856\times\frac{45.6}{8.95}\times265\). Let's calculate it precisely: \(\frac{45.6\times0.856\times265}{8.95}=\frac{45.6\times227.84}{8.95}=\frac{10399.504}{8.95}\approx1162\space N\) (there might be slight difference due to rounding in MA earlier).

Answer:

The IMA of the wheel and axle is approximately \(5.10\) (or more precisely \(\frac{45.6}{8.95}\approx5.095\)).

Part b: MA of Wheel and Axle