# tension in wires from which a block is suspended

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I am trying to figure out why the answer is T>W. Do you have any idea why? If so could you please explain. Thank you.

edited Jan 6
The tension force will be larger since it has to balance the mass weight which is downward and since the system is in equilibrium the tension force will be a bit larger

In order to compare $T$ and $W$, one needs to express one in terms of the other. As explained in the diagram itself, since the block is in translational equilibrium, we have:

$$\vec{T_1}+\vec{T_2}=\vec{W_b}$$

Projecting this relation onto the $Oy$ axis, the components of the tensions are both $T\sin\theta$ (because the angle between $T$ and $Oy$ is $\frac{\pi}{2}-\theta$ and the angle between $T$ and $Ox$ is $\theta$). Therefore:

$$2T\sin\theta=W\implies T=\dfrac{W}{2\sin\theta}$$

The information the strings are almost horizontal is equivalent to $\theta\cong 0$. For very small angles, the sine function is as well extremely small, and almost equal to the angle $\theta$ itself. Take a look at the following examples (I've included them here solely to make sure that you understand why $\sin\theta\to 0$ as $\theta\to 0$)

• $\sin(0.01)\cong 0.00999983333333\cong 0.01$
• $\sin(0.0001) \cong 0.00009999999983\cong 0.0001$
• $\sin(0)=0$

So the denominator $2\sin\theta\cong0$ and therefore $T\to\infty$. As $W=mg$, clearly $W$ is a finite value (making the reasonable assumption that the mass is not inifite), and therefore clearly $T>W$.

answered Jan 11 by (120 points)