Question

A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above the horizontal, and pull. You find that the minimum tension you can apply to the rope in order to make the box start moving is Tmin. Find the coefficient of static friction between the floor and the box.

Answer 1

Answer:

The correct answer to the following question will be "$\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$".

Explanation:

According to the question,

$\sum F_{x}$

⇒  $TCos \theta-F_{s}=0$

⇒  $T_{m}Cos \theta =F_{s}$ ...(equation 1)

$\sum F_{y}$

⇒  $TSin \theta+F_{N}=m_{g}$

⇒  $M_{g}-TSin \theta=F_{N}$ ...(equation 2)

Now,

From equation 1 and equation 2, we get

⇒  $T_{m} Cos \theta = \mu_{s}F_{N}$

On putting the value of $F_{N}$, we get

⇒  $T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)$

⇒  $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$