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A (lightweight) pallet has a load of ten identical cases of tomato paste (see Fig. 7–39), each of which is a cube of length \(l\). Find the center of gravity in the horizontal plane, so that the crane operator can pick up the load without tipping it.

FIGURE 7-39

Problem 53.

Short Answer

Expert verified

The center of mass of the mass system is at \(\left( {1.2l,0.9l} \right)\) point relative to the origin at the upper left corner.

Step by step solution

01

Given data

To find the position of the CM in two dimensions, you have to find both the x and y positions of the CM.To find the center of mass, you have to find the x and y coordinates of the CM for individual cases.

The sides of each cube are l.

You can assume the mass of the cubes to be m as they are the same in size.

Now, the top view of the pallet is shown below:


Here, you can also consider the upper left corner as the origin.

02

Calculation of the center of mass

As the cases are symmetrical in shape, the CM of an individual box is at the center of the box.

Therefore, the x coordinate of the first column masses \(\left( {3m,m,m} \right)\) is \(\frac{l}{2}\).

The x coordinate for the second column masses \(\left( {2m,m} \right)\) is \(\frac{{3l}}{2}\).

The x coordinate for the last column mass \(2m\) is \(\frac{{5l}}{2}\).

The y coordinate for the first row masses \(\left( {3m,2m,2m} \right)\) is \(\frac{l}{2}\).

The y coordinate for the second row masses \(\left( {m,m} \right)\) is \(\frac{{3l}}{2}\).

The y coordinate for the third row mass \(\left( m \right)\) is \(\frac{{5l}}{2}\).

Now, the x coordinate of the center of mass of the cases is:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{\left\{ {\left( {3m + m + m} \right) \times \frac{l}{2}} \right\} + \left\{ {\left( {2m + m} \right) \times \frac{{3l}}{2}} \right\} + \left( {2m \times \frac{{5l}}{2}} \right)}}{{\left( {3m + m + m} \right) + \left( {2m + m} \right) + 2m}}\\ = \frac{{\frac{{5ml}}{2} + \frac{{9ml}}{2} + \frac{{10ml}}{2}}}{{10m}}\\ = 1.2l\end{array}\)

Now, the y coordinate of the center of mass of the cases is:

\(\begin{array}{c}{y_{{\rm{CM}}}} = \frac{{\left\{ {\left( {3m + 2m + 2m} \right) \times \frac{l}{2}} \right\} + \left\{ {\left( {m + m} \right) \times \frac{{3l}}{2}} \right\} + \left( {m \times \frac{{5l}}{2}} \right)}}{{\left( {3m + m + m} \right) + \left( {2m + m} \right) + 2m}}\\ = \frac{{\frac{{7ml}}{2} + \frac{{6ml}}{2} + \frac{{5ml}}{2}}}{{10m}}\\ = 0.9l\end{array}\)

Hence, the center of mass of the mass system is at \(\left( {1.2l,0.9l} \right)\) point relative to the origin at the upper left corner.

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Most popular questions from this chapter

When a high jumper is in a position such that his arms and lower legs are hanging vertically, and his thighs, trunk, and head are horizontal just above the bar, estimate how far below the torso’s median line the CM will be. Will this CM be outside the body? Use Table 7–1.

Billiard balls A and B, of equal mass, move at right angles and meet at the origin of an xy coordinate system as shown in Fig. 7–36. Initially ball A is moving along the y axis at \({\bf{ + 2}}{\bf{.0}}\;{\bf{m/s}}\), and ball B is moving to the right along the x axis with speed \({\bf{ + 3}}{\bf{.7}}\;{\bf{m/s}}\).After the collision (assumed elastic), ball B is moving along the positive y axis (Fig. 7–36) with velocity What is the final direction of ball A, and what are the speeds of the two balls?

FIGURE 7-36

Problem 46. (Ball A after the Collison is not shown)

An atomic nucleus of mass m traveling with speed v collides elastically with a target particle of mass 2m (initially at rest) and is scattered at 90°.

(a) At what angle does the target particle move after the collision?

(b) What are the final speeds of the two particles?

(c) What fraction of the initial kinetic energy is transferred to the target particle?

A neon atom \(\left( {m = 20.0\;{\rm{u}}} \right)\) makes a perfectly elastic collision with another atom at rest. After the impact, the neon atom travels away at a 55.6° angle from its original direction and the unknown atom travels away at a \( - {50.0^ \circ }\) angle. What is the mass (in u) of the unknown atom? [Hint: You could use the law of sines.]

A huge balloon and its gondola, of mass M, are in the air and stationary with respect to the ground. A passenger, of mass m, then climbs out and slides down a rope with speed v, measured with respect to the balloon. With what speed and direction (relative to Earth) does the balloon then move? What happens if the passenger stops?

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