Question

tension T in the string as shown in the figure is

Answer 1

Answer: The tension in the string is $T = (49-34.3\sqrt3)N$

Explanation:

Given that,

Mass m = 10 kg

Angle $\theta = 30^{0}$

Coefficient $\mh = 0.7$

According to figure,

$T + f_{\mu} = mg sin 30^{0}$

$T + \mu mg\cos 30^{0} = mg\sin 30^{0}$

$T = 10\kg\times 9.8m/s^{2}\times\dfrac{1}{2}- 0.7\times 10\ kg\times9.8m/s^{2}\times\dfrac{\sqrt3}{2}$

$T = (49-34.3\sqrt3)\ N$

Hence, the tension in the string is $T = (49-34.3\sqrt3)N$.

Answer 2

Answer:

The measured tension in the string is $\bold{T =(49-34.3\sqrt { 3 } )N}$

Explanation:

From the figure it is inferred that mass of the string m= 10 kg  

Angle mentioned here is $\theta \quad =\quad { 30 }^{ \circ}$

The value of coefficient is 0.7

From the figure, it is clear that,

$T+{ F }_{ \mu}\quad =\quad mg\sin { { 30 }^{ \circ}}$

$T+\mu mg\cos { { 30 }^{ \circ}} \quad =\quad mg\sin { { 30 }^{ \circ} }$

$T\quad =\quad 10\times 9.8\frac { m }{ { s }^{ 2 } } \times \frac { 1 }{ 2 } -0.7\times 10Kg\times 9.8\frac { m }{ { s }^{ 2 } }$

$T\quad =\quad (49-34.3)N$

Therefore the measured tension in the string is $\bold{T =(49-34.3\sqrt { 3 } )N}$