91Ó°ÊÓ

In the given figure, the sled dog in drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10. If the tension in rope 1 is 150 N,

Consider the figure with direction of forces acting as :

Further, both A & B are moving together their accelerations will be the same. Let acceleration be {a}.

Consider the motion of sled A :

For this, consider different forces acting with their directions as

(1) Weight of A : $m_{1}g=100(9.8)=980N.$(downward)

(2) $N_1$is normal force of earth and is equal to its weight in upward.

$⇒N_1=980\mathrm{N}.$(upward)

(3) The tension of the rope $1:T_1=150\mathrm{N}$(forward), and

(4) The friction force $=F_{f1}=\mathrm{μ}N1=(0.10)\left(980\right)=98\mathrm{N}$(backward)

$∴$Equation of motion for A is

$T_1-\mathrm{μ}N1=m_1\mathrm{a}⋯\left(1\right)$

Substituting the values of given data in equation (1)

$⇒a=\frac{(150-98)}{100}=0.52\mathrm{m}/{\mathrm{s}}^2⋯\left(2\right)$

For motion of sled B : ---

consider different forces acting with their directions as

(1) Weight of the sled : $m_{2}g=80kg×9.8m/s^2=784\mathrm{N}$(downward)

(2) ${\mathrm{N}}_2$is normal force of earth and is equal to its weight in upward.

$⇒N_1=784\mathrm{N}$(upward)

(3) The tension of the rope localid="1649264946519" $2:T_1=?\text{(forward),}$(forward),

(4) The friction force $=F_{f2}=\mathrm{μ}{N}_2=(0.10)\left(784\right)=78.4\mathrm{N}$(backward) and

(5) The tension of rope $1:{\mathrm{T}}_1=150\mathrm{N}$(backward)

Equation of motion for B is $T_2-\mathrm{μ}{N}_2-T_1=m_{2}a⋯\left(3\right)$

Substituting the values of given data in equation (3)

localid="1650275764639" $$\begin{gathered} ⇒T_2-78.4-150=80×(0.52) \\ ⇒T_2=270\mathrm{N} \end{gathered}$$

The tension in rope$2\text{is}270N.$