Why would tension be smaller if the string were parallel to the lab bench? May 30, 2018

Let $M$ be mass of block and $m$ be mass suspended with an inextensible string, $\mu$ be coefficient of friction, $\theta$ be angle made by string with the horizontal where $\theta \ge 0$ and $T$ be tension, (reaction force) in the strings. It is given that block has a movement. Let $a$ be its acceleration. As both masses are joined with a common string, the hanging mass also moves downwards with the same acceleration.
Taking East as positive $x$-axis and North as positive $y$-axis.

External forces responsible for the magnitude of acceleration of masses when considered as single object

$\left(M + m\right) a = m g \cos \theta - \mu \left(M g - m g \sin \theta\right)$ ......(1)

For Block it is $x$ component of tension which is responsible for its acceleration.

$a = {T}_{x} / M$
$\implies a = \frac{T \cos \theta}{M}$
$\implies T = \frac{M a}{\cos} \theta$
$\implies T = \frac{M \left(m g \cos \theta - \mu \left(M g - m g \sin \theta\right)\right)}{\left(M + m\right) \cos \theta}$ .....(2)

Rewriting it as

$T = a - \frac{b}{\cos} \theta + c \tan \theta$
where $a , b \mathmr{and} c$ are system parameters defined with help of (2) not dependent on $\theta$

We see that $T$ is dependent on two terms involving $\theta$

1. $- \frac{1}{\cos} \theta$. For $T$ to be a smaller number $\cos \theta$ term must be maximum. We know that $\cos \theta$ has a maximum value $= 1$ for $\theta = {0}^{\circ}$
2. $\tan \theta$. For $T$ to be a smaller number, $\tan \theta$ term must be zero. We know that $\tan \theta$ has a value $= 0$ for $\theta = {0}^{\circ}$.

Hence, we see that tension will be smaller if the string connecting the block were parallel to the lab bench.