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

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

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

Answer 2

The coefficient of static friction between the floor and the box is $\mu_s = \frac{T_mCos\theta}{M_9-T_mSin\theta^n}$

Coefficient of static friction:

Static friction should be the force where an object should be at rest. It is the friction where the individuals try to shift a stationary object on a surface, without triggering any motion that lies between the body and the surface.

Since

$\sum F_x$ should be

T Cosθ  - F_s =

Now

T_mCosθ = F_s ..........(equation 1)

Now

$\sum F_y$

T Sinθ + F_N = m_g

M_g - TSinθ  = F_N............(equation 2)

Now

T_mCosθ  = \mu_s F_N

Now

T_mCos\mu = \mu(M_g - T_mSinθ)

So, $\mu_s = \frac{T_mCos\theta}{M_9-T_mSin\theta^n}$

Hence, The coefficient of static friction between the floor and the box is $\mu_s = \frac{T_mCos\theta}{M_9-T_mSin\theta^n}$

Learn more about friction here: https://brainly.com/question/18519949