A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force F. If the rope AB makes an angle θ with the vertical in equilibrium, then the tension in the string AB is:
Let us see the Free body Diagram of the arrangement
For Equilibrium, T Cos θ = Mg and T Sin θ = F ;
As, T Sin θ = F ; T= F/ Sinθ
Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A 15 kg weight is attached to the rope at the mid-point which now no longer remains horizontal. The minimum tension required to completely straighten the rope is:
For Equilibrium condition,
2T cos θ = mg ⇒ T = mg/(2 cos θ)
As, we try to make string horizontal θ approaches 90°. so Tension will become T= mg/2 cos (90°)
i.e T becomes infinite
A flexible chain of weight W hangs between two fixed points A and B at the same level. The inclination of the chain with the horizontal at the two points of support is θ. What is the tension of the chain at the endpoint?
For Equilibrium in vertical direction
2T sin θ = W ⇒ T = W/(2 sin θ) = (W cosec θ) / 2
A body of weight W1 is suspended from the ceiling of a room through a chain of weight W2. The ceiling pulls the chain by a force
A block of mass 10 kg is suspended through two light spring balances as shown in figure
Always remember that spring balance measure Tension in the string in form of kg-wt. As tension in the string is 10 kg-wt.
Reading of Both the scales will read 10 kg
Block A of mass 4 kg is to be kept at rest against a smooth vertical wall by applying a force F as shown in figure. The force required is (g = 10 m/s²)
In the given arrangement, the normal force applied by block on the ground is
For Equilibrium in vertical direction
F cos θ + N = mg
or N = mg - F cosθ
The value of \( \frac{T_{3}}{T_{1}} \) is
\[ \]
As the arrangement is in equilibrium T3= 150 N and T1 = 50 N
so T3/T1=3
The ratio of tension T1 and T2 is (strings are massless)
For Equilibrium Condition,
T1 = 70 N and T2 = 50 N so, T1: T2 = 7 : 5
A man of mass m is standing on a board and pulling the board of mass m up with force F by the pulley system as shown. Normal reaction between man and board is
As the person is in equilibrium net force acting on it should be zero,
so, mg = N + F
⇒ N = mg - F