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

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1 Answer
May 30, 2018

Let MM be mass of block and mm be mass suspended with an inextensible string, muμ be coefficient of friction, thetaθ be angle made by string with the horizontal where theta>=0θ0 and TT be tension, (reaction force) in the strings. It is given that block has a movement. Let aa 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 xx-axis and North as positive yy-axis.

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

(M+m)a=mgcostheta-mu(Mg-mgsintheta)(M+m)a=mgcosθμ(Mgmgsinθ) ......(1)

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

a=T_x/Ma=TxM
=>a=(Tcostheta)/Ma=TcosθM
=>T=(Ma)/costhetaT=Macosθ
=>T=(M(mgcostheta-mu(Mg-mgsintheta)))/((M+m)costheta)T=M(mgcosθμ(Mgmgsinθ))(M+m)cosθ .....(2)

Rewriting it as

T=a-b/costheta+ctanthetaT=abcosθ+ctanθ
where a,b and ca,bandc are system parameters defined with help of (2) not dependent on thetaθ

We see that TT is dependent on two terms involving thetaθ

  1. -1/costheta1cosθ. For TT to be a smaller number costhetacosθ term must be maximum. We know that costhetacosθ has a maximum value =1=1 for theta=0^@θ=0
  2. tanthetatanθ. For TT to be a smaller number, tanthetatanθ term must be zero. We know that tanthetatanθ has a value =0=0 for theta=0^@θ=0.

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