Answered

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I have some problems applying the formulas to solve physics problems. I understand all the concepts needed, but just freeze when I see questions, especially when it comes to trying to combine linear and rotational conceptsA uniform, 255 N rod that is 1.90 m long carries a 225 N weight at its right end and an unknown weight W toward the left end (Figure 1). When W is placed 60.0 cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 75.0 cm from the right end.1) Find W.2) If W is now moved 30.0 cm to the right, how far must the fulcrum be moved to restore balance?

I Have Some Problems Applying The Formulas To Solve Physics Problems I Understand All The Concepts Needed But Just Freeze When I See Questions Especially When I class=

Sagot :

ANSWER

[tex]\begin{gathered} 1)\text{ }214.90\text{ }N \\ \\ 2)\text{ }0.09\text{ }m \end{gathered}[/tex]

EXPLANATION

First, let us make a sketch of the diagram showing the distances on the rod:

1) Since the fulcrum is balanced, the center of gravity of the system will be at the fulcrum.

The center of gravity (in the horizontal is given by:

[tex]x=\frac{W_1x_1+W_2x_2+W_3x_3}{W_1+W_2+W_3}[/tex]

where W1 = the weight on the right end = 225 N

W2 = the weight of the rod = 255 N

W3 = the weight place on the left = W

x1 = the position of W1 (taking the left as the origin) = 1.90 m

x2 = the position of the center of mass of the rod = x1/2 = 0.95 m

x3 = the position of W from the left end = 0.60 m

x = position of center of gravity of the rod from the left end i.e. at the fulcrum = 1.90 - 0.75 = 1.15 m

Now, substitute the values given in the question and solve for W:

[tex]\begin{gathered} 1.15=\frac{(225*1.90)+(255*0.95)+(W*0.60)}{225+255+W} \\ \\ 1.15=\frac{427.5+242.25+0.60W}{480+W} \\ \\ 1.15(480+W)=669.75+0.60W \\ \\ 552+1.15W=669.75+0.60W \\ \\ 1.15W-0.60W=669.75-552 \\ \\ 0.55W=117.75 \\ \\ W=\frac{117.75}{0.55} \\ \\ W=214.09\text{ }N \end{gathered}[/tex]

That is the value of W.

2) Now, W is moved 30.0 cm (0.30 m) to the right.

This implies that:

[tex]x_3=0.60+0.30=0.90\text{ }m[/tex]

Since the other values (including W) do not change, we can now solve for x, which is the new center of gravity:

[tex]\begin{gathered} x=\frac{(225\times1.90)+(255\times0.95)+(214.09\times0.90)}{225+255+214.09} \\ \\ x=\frac{427.5+242.25+192.681}{694.09}=\frac{862.431}{694.09} \\ \\ x=1.24\text{ }m \end{gathered}[/tex]

Therefore, the fulcrum must be moved:

[tex]\begin{gathered} 1.24\text{ }m-1.15\text{ }m \\ \\ 0.09\text{ }m \end{gathered}[/tex]

The fulcrum should be moved 0.09 m to the right (since the W is moved to the right).

View image SimranD574936
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