Question 88: Laws of Motion

Question

As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g = 10 m/s$$^2$$)

Accepted Answer

Let's consider $T$ tension in string and a acceleration of blocks.
FBD of $4 \mathrm{~kg}$
$4 g \sin 	heta-\mathrm{T}=4 \mathrm{a}$
$$
\frac{4 g \sqrt{3}}{2}-T=4 a
$$ .........(1)
FBD of $1 \mathrm{~kg}$
$T-g \sin 30^{\circ}=a$
$T-\frac{g}{2}=a \quad\quad...(2)$
From (1) and (2), we get
$${{4g\sqrt 3 } \over 2} - {g \over 2} = 5a$$
$$ \Rightarrow $$ $${{g\left( {4\sqrt 3  - 1} ight)} \over 2} = 5a$$
$$ \Rightarrow $$ $$5a = {{10\left( {4\sqrt 3  - 1} ight)} \over 2}$$
$$ \Rightarrow $$ $$a = \left( {4\sqrt 3  - 1} ight)$$ m/s
2
By putting value of "a" in equation (2), we get
$$T - {{10} \over 2} = \left( {4\sqrt 3  - 1} ight)$$
$$ \Rightarrow $$ $$T = \left( {4\sqrt 3  - 1} ight) + 5$$ = $$\left( {4\sqrt 3  + 4} ight)$$
$$ \Rightarrow $$ $T=4(\sqrt{3} + 1) N$

Answer

$$\left( {4\sqrt 3  - 1} \right)N$$

Answer

$$\left( {4\sqrt 3  + 1} \right)N$$

Answer

$$4\left( {\sqrt 3  - 1} \right)N$$

Answer

$$4\left( {\sqrt 3  + 1} \right)N$$

Laws of Motion

As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g = 10 m/s$^2$)

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