# tension in wires from which a block is suspended

–1 vote
39 views 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.

asked Jan 6 1 flag
edited Jan 6
There is an explanation in the diagram. What  don't you like about this explanation?
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)
This answer is already provided in the question.