Chapter 1: Q87P (page 366)

You need to measure the mass Mof a 4.00 mlong bar. The bar has a square cross section but has some holes drilled along its length, so you suspect that its center of gravity isn’t in the middle of the bar. The bar is too long for you to weigh on your scale. So, first, you balance the bar on a knife-edge pivot and determine that the bar’s center of gravity is 1.88 mfrom its left-hand end. You then place the bar on the pivot so that the point of support is 1.50 mfrom the left-hand end of the bar. Next you suspend amass $m_{2} = 1.00 kg$ from the bar at a point from the left-hand end. Finally, you suspend a massfrom the bar at a distance xfrom the left-hand end and adjustso that the bar is balanced. You repeat this step for other values ofand record each corresponding value of. The table gives your result
Draw a free-body diagram for the bar whenandare suspended from it.
Apply the static equilibrium equation $\sum_{} τ_{z} = 0$ with the axis at the location of the knife-edge pivot. Solve the equation for xas a function of $m_{2}$ .
Plot xversus $\frac{1}{m_{2}}$ . Use the slope of the best-fit straight line and the equation you derived in part (b) to calculate that bar’s mass M. Use $g = 9.80 m / s^{2}$ .
What is the y-intercept of the straight line that fits the data? Explain why it has this value

Short Answer

Expert verified

1.Free body diagram for bar is given by,

2. $x = 1.50 + \frac{(1.30) m_{1} - (0.38 m) M}{m_{2}}$

3. Graph of x versus $\frac{1}{m_{2}}$ is given by,

Mass of bar is M = 1.607 kg .

y-intercept is at x = 1.50 m

Step by step solution

Static equilibrium

Any system that is in a state of static equilibrium has all of its forces and torques added together to equal zero.

Identification of given data

Here we have given that, mass of the bar is M

$m_{1} = 2.00 kg And m_{2} = 1.00 kg$

Length of bar is L = 4.00 m

Step 3: Draw the free body diagram.

(a)

Free body diagram for bar is given by,

Solving the equation for x as a function of m2 by applying the static equilibrium equation ∑ τz=0 with the axis at the location of the knife-edge pivot.

(b)

Now, for static equilibrium, $\sum_{} τ_{A} = 0$

$\begin{aligned} (1.50 m - 0.200 m) m_{1} g - (1.88 m - 1.50 m) M g - (x - 1.50 m) m_{2} g = 0 \\ x = 1.50 + \frac{(1.30 m) m_{1} - (0.38 m) M}{m_{2}} \end{aligned}$

Plot x versus 1m2 .calculate that bar’s mass M Use the slope of the best-fit straight line and the equation you derived in part (b)

(c)

From the given data table of x and $\frac{1}{m_{2}}$ is given by,

Now, from the above table graph of x versus $\frac{1}{m_{2}}$ is given by,

Now, from graph, best fit line is given by,

$\begin{aligned} x = 1 .9895 (\frac{1}{m_{2}}) + 1 .5072 (1) \\ Also, from step 4 we have, \\ x = 1 .50 + \frac{(1 .30 m) m_{1} - (0 .38 m) M}{m_{2}} \\ x = ((1 .30 m) m_{1} - (0 .38 m) M) \frac{1}{m_{2}} + 1 .50 (2) \end{aligned}$

Now, from equation (1) and (2) we have,

$\begin{aligned} (1.30 m) m_{1} - (0.38 m) M = 1.9895 \\ (1.30 m) (2.00 kg) - (0.38 m) M = 1.9895 \\ M = \frac{1.9895 - (1.30 m) (2.00 kg)}{(0.38 m)} \\ M = 1.607 kg \end{aligned}$