Why are mountain roads generally made winding upwards rather than going straight up?
While going up a mountain, the force of friction acting on a vehicle of mass $$m$$ is $$f=\mu R=\mu \mathrm{mg} \cos \theta$$, where $$\theta$$ is the angle of slope of the road with the horizontal. To avoid skidding force of friction ($$f$$) should be large and therefore, $$\cos \theta$$ should be large and hence, $$\theta$$ should be small.
That's why mountain roads are generally made winding upwards rather than going straight upto avoid skidding.
A mass of 2 kg is suspended with thread $$A B$$ (figure). Thread $$C D$$ of the same type is attached to the other end of 2 kg mass. Lower thread is pulled gradually, harder and harder in the downward direction, so as to apply force on $$A B$$. Which of the threads will break and why?
The thread $$A B$$ will break earlier than the thread $$C D$$. This is because force acting on thread $$C D=$$ applied force and force acting on thread $$A B=$$ (applied force + weight of 2 kg mass). Hence, force acting on thread $$A B$$ is larger than the force acting on thread $$C D$$.
In the above given problem if the lower thread is pulled with a jerk, what happens?
When the lower thread $$C D$$ is pulled with a jerk, the thread $$C D$$ itself break. Because pull on thread $$C D$$ is not transmitted to the thread $$A B$$ instantly.
Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in figure. Calculate $$T_1$$ and $$T_2$$ when whole system is going upwards with acceleration $$=2 \mathrm{~m} / \mathrm{s}^2$$ (use $$\mathrm{g}=9.8 \mathrm{~ms}^{-2}$$).
$$\begin{aligned} \text { Given, } m_1 & =5 \mathrm{~kg}, m_2=3 \mathrm{~kg} \\ g & =9.8 \mathrm{~m} / \mathrm{s}^2 \text { and } a=2 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$$
For the upper block
$$\begin{aligned} & & T_1-T_2-5 g & =5 a \\ \Rightarrow & & T_1-T_2 & =5(g+a) \end{aligned}$$
For the lower block
$$T_2-3 g=3 a$$
$$\Rightarrow \quad T_2=3(g+a)=3(9.8+2)=35.4 \mathrm{~N}$$
$$\begin{aligned} &\text { From Eq. (i) }\\ &\begin{aligned} T_1 & =T_2+5(g+a) \\ & =35.4+5(9.8+2)=94.4 \mathrm{~N} \end{aligned} \end{aligned}$$
Block $$A$$ of weight 100 N rests on a frictionless inclined plane of slope angle $$30^{\circ}$$. A flexible cord attached to $$A$$ passes over a frictionless pulley and is connected to block $$B$$ of weight $$w$$. Find the weight $$w$$ for which the system is in equilibrium.
In equilibrium, the force $$m g \sin \theta$$ acting on block $$A$$ parallel to the plane should be balanced by the tension in the string, i.e.,
$$\begin{aligned} m g \sin \theta & =T=F \quad [\because \text{T=F given}]\quad \text{.... (i)}\\ w & =T=F \quad \text{... (ii)} \end{aligned}$$
and for block B, where, $$w$$ is the weight of block $$B$$.
From Eqs. (i) and (ii), we get,
$$\begin{aligned} \therefore \quad W & =m g \sin \theta \\ & =100 \times \sin 30^{\circ} \quad (\because m g=100 \mathrm{~N})\\ & =100 \times \frac{1}{2} \mathrm{~N}=50 \mathrm{~N} \end{aligned}$$