Laws of Motion
Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are frictionless. Find the tension in the string.
Two blocks, of weights $$3.6$$ N and $$7.2$$ N, are connected by a massless string and slide down a $$30^\circ$$ inclined plane. The coefficient of kinetic friction between the lighter block and the plane is $$0.10$$, and the coefficient between the heavier block and the plane is $$0.20$$. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the taut string
The free-body diagrams for the two blocks are shown below. T is the magnitude of
the tension force of the string, $$\vec F_{NA}$$ is the normal force on block A (the leading block), $$\vec F_{NB}$$ is the normal force on block B, $$\vec f_{A}$$ is kinetic friction force on block A, $$\vec f_{B}$$ is kinetic
friction force on block B. Also, $$m_A$$ is the mass of block A (where $$m_A = W_A/g$$ and $$W_A = 3.6$$ N), and $$m_B$$ is the mass of block B (where $$m_B = W_B/g$$ and $$W_B = 7.2$$ N). The angle of the
incline is $$\theta = 30^\circ$$.
For each block, we take +x downhill (which is toward the lower-left in these diagrams)
and +y in the direction of the normal force. Applying Newton’s second law to the x and y
directions of both blocks A and B, we arrive at four equations:
$$W_A\sin\theta-f_A-T m_Aa$$
$$ F_{NA}-W_Acos\theta=0$$
$$W_B\sin\theta-f_B+T m_Bb$$
$$ F_{NB}-W_Bcos\theta=0$$
which, when combined with Eq. ($$ f_A =\mu_{KA}F_{NA} $$ where $$\mu_{KA} = 0.10$$ and$$ f_B =\mu_{KB}F_{NB}f_B $$ where $$\mu_{KB} = 0.20$$), fully describe the dynamics of the system so long as the blocks have
the same acceleration and $$T > 0$$.
(a) From these equations, we find the acceleration to be
$$a=g\Bigg(\sin\theta-\dfrac{\mu_{KA}W_A+\mu_{KB}W_B}{W_A+W_B}cos\theta\Bigg)=3.5m/s^2$$
(b) We solve the above equations for the tension and obtain
$$T=\dfrac{W_AW_B}{W_A+W_B}(\mu_{KA}-\mu_{KB})cos\theta=0.21N$$
Note: The tension in the string is proportional to $$\mu_{KA}-\mu_{KB}$$, the difference in coefficients
of kinetic friction for the two blocks. When the coefficients are equal ($$ \mu_{KA}=\mu_{KB} $$), the
two blocks can be viewed as moving independently of one another and the tension is zero.
Similarly, when $$\mu_{KA}<\mu_{KB}$$ (the leading block A has larger coefficient than the B), the string is slack, so the tension is also zero.
Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are frictionless. Find the tension in the string.
In the given arrangement, $$n$$ number of equal masses are connected by strings of negligible masses. The tension in the string connected to $$n$$th mass is -
Three solids of mass $$m_{1}$$ , $$m_{2}$$ and $$m_{3}$$ are connected with weight less string in succession and are placed on a frictionless table. If the mass $$m_{3}$$ is dragged with a force $$T$$. The tension in the string between $$m_{2}$$ and $$m_{3}$$ is:-
A man is pulling a rope attached to a block on a smooth horizontal table. The tension in the rope will be the same at all points.
Which of the following expressions correctly represents $$T_{1}$$ and $$T_{2}$$ if the system is given an upward acceleration by a pulling up mass $$A$$?
A uniform fine chain of length $$l$$ is suspended with lower end just touching a horizontal table. The pressure on the table, when a length $$x$$ has reached the table is
A uniform chain is coiled up on a horizontal plane and one end passes over a small light pulley at a height $$'a'$$ above the plane. Initially, a length $$'b'$$ hangs freely on the other side. If $$b = 2a$$.
A flat car is given ar acceleration $$a_{0} = 2\ m/s^{2}$$ starting from rest. A cable is connected to a crate $$A$$ of weight $$50\ kg$$ as shown. Neglect friction between the floor and the car wheels and also the mass of the pulley. Calculate corresponding tension in the cable if $$\mu = 0.30$$ between the crate and the floor of the car.
Figure shows a block of mass $$m$$ kept on inclined plane with inclination $$\theta$$. The tension in the string is
Three blocks of masses $$m_{1}, m_{2}$$ and $$m_{3}$$ are placed on a horizontal frictionless surface. A force of $$40\ N$$ pulls the system then calculate the value of $$T$$, if $$m_{1} = 10\ kg, m_{2} = 6\ kg, m_{3} = 4\ kg$$.