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17. A uniform rod of length 120cm and mass 0.7kg, is suspended horizontally by two vertical strings at points 12cm from each end. If masses of 4.0 and 2.0kg are hung on it at points 20cm respectively from one end and 25cm. from the other, what would the tension in the strings be?

pls answer quickly​

User Freginold
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Answer:

To solve this problem, you can use the principle of transverse static equilibrium. This states that the sum of the forces acting on an object must be equal to zero for the object to be in equilibrium. In this case, the forces acting on the rod are the tension in the strings, the weight of the rod, and the weight of the masses hung on the rod.

First, you can find the weight of the rod by multiplying its mass by the acceleration due to gravity:

Wrod = mrod * g

= 0.7 kg * 9.8 m/s^2

= 6.86 N

Next, you can find the weight of the 4.0-kg mass by multiplying its mass by the acceleration due to gravity:

W1 = m1 * g

= 4.0 kg * 9.8 m/s^2

= 39.2 N

You can find the weight of the 2.0-kg mass in the same way:

W2 = m2 * g

= 2.0 kg * 9.8 m/s^2

= 19.6 N

To find the tension in the strings, you can use the principle of transverse static equilibrium. The forces acting on the rod are the tension in the strings, the weight of the rod, and the weight of the masses hung on the rod. You can represent these forces as vectors acting at the points where the strings are attached to the rod. The vectors representing the tension in the strings will be equal in magnitude but opposite in direction. The vectors representing the weight of the rod and the masses will act vertically downward.

[asy]

unitsize(2cm);

pair P1, P2, P3, P4;

P1 = (0,0);

P2 = (1,0);

P3 = (0.8,0);

P4 = (0.2,0);

draw((-1,0)--(2,0));

draw((0,-1)--(0,1));

draw(P1--P3,red,Arrow(6));

draw(P2--P4,red,Arrow(6));

draw(P3--P4,red,Arrow(6));

draw((P1)--(P1+(0,-0.7)));

draw((P2)--(P2+(0,-4)));

draw((P2)--(P2+(0,-2)));

label("$T_1$", (P1 + P3)/2, red);

label("$T_2$", (P2 + P4)/2, red);

label("$W_{rod}$", (P3 + P4)/2, red);

label("$W_1$", (P2 + P4 + (0,-4))/2, red);

label("$W_2$", (P2 + P4 + (0,-2))/2, red);

label("$12$ cm", (0,0.2), red);

label("$20$ cm", (0.4,-0.4), red);

label("$25$ cm", (0.75,-0.4), red);

[/asy]

The sum of the forces in the x-direction must be equal to zero, so you can set up the following equation:

T1 - T2 = 0

The sum of the forces in the y-direction must also be equal to zero, so you can

User Wei WANG
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