Why would tension be smaller if the string were parallel to the lab bench?

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1 Answer
May 30, 2018

Let M be mass of block and m be mass suspended with an inextensible string, μ be coefficient of friction, θ be angle made by string with the horizontal where θ0 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

(M+m)a=mgcosθμ(Mgmgsinθ) ......(1)

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

a=TxM
a=TcosθM
T=Macosθ
T=M(mgcosθμ(Mgmgsinθ))(M+m)cosθ .....(2)

Rewriting it as

T=abcosθ+ctanθ
where a,bandc are system parameters defined with help of (2) not dependent on θ

We see that T is dependent on two terms involving θ

  1. 1cosθ. For T to be a smaller number cosθ term must be maximum. We know that cosθ has a maximum value =1 for θ=0
  2. tanθ. For T to be a smaller number, tanθ term must be zero. We know that tanθ has a value =0 for θ=0.

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