You have a 10-kg table with each leg of mass 1.0 kg total mass 14 kg. If you pla

Question

You have a 10-kg table with each leg of mass 1.0 kg total mass 14 kg. If you place a 5.0-kg pot of soup in the back right corner of the table (Figure 1) , where is the table's center of mass? Assume that the size of the soup bowl and thickness of the tabletop are negligible.

Part A

Determine the x-coordinate of the table's center of mass.

Express your answer in terms of L.

Part B

Determine the y-coordinate of the table's center of mass.

Express your answer in terms of w.

Part C

Determine the z-coordinate of the table's center of mass.

Express your answer in terms of h.

Anwers:

A. 5L/38

B.5w/38

c. (2/19)h

How do I find these answers?

Please write out step by step on how to do the problem

Explanation / Answer

A.

Note that

xCOM = Sum (mi xi) / Sum (mi)

Here,

m1 = 10 kg, x1 = 0
For the two left legs, m2 = m3 = 1.0 kg, x2 = x3 = -L/2
For the two right legs, m4 = m5 = 1.0 kg, x4 = x5 = -L/2
For the pot, m6 = 5.0 kg, x6 = L/2

Thus, using that formula,

xCOM = 5L/38 [ANSWER]

**********************

Note that

yCOM = Sum (mi yi) / Sum (mi)

By symmetry, the yCOM without the pot is at yCOM = 0.

Thus, only the pot contributes on the numerator,

ypot = w/2

Thus,

yCOM = 5w/38 [ANSWER]

***************************

Note that

ZCOM = Sum (mi Zi) / Sum (mi)

Here,

m1 = 10 kg, Z1 = 0
For the legs, m2 = m3 = m4 = m5 1.0 kg, z2 = z3 = z4 = z5 = -h/2
For the pot, m6 = 5.0 kg, z6 = 0

Thus, using the formula and simplifying,

zCOM = 2h/19 [ANSWER]

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