An external force F is holding a bob of mass 725 g in a stationary position. The angle that the massless rope makes with the vertical is θ = 33°.
(a) What is the magnitude, F, of the force needed to maintain equilibrium in N?
(b) What is the tension in the rope in N?
F =T•sinα ,
mg = T•cos α..
F/m•g = tan α,
F = m•g•tan α =0.725•9.8•0.65 =4.6 N,
T =F/sinα = 4.6/0.54 = 8.45 N.
I understand how the first and third equations are used to find the answers, but how is the second equation (mg = Tcosa) used?
The third equation is the result of division the first equation by the second equation
To solve these problems, we can use Newton's laws of motion. Let's break down each part:
(a) To find the magnitude of the force needed to maintain equilibrium (F), we need to consider the forces acting on the bob. In this case, there are two forces: the force of gravity (weight) and the tension in the rope.
The force of gravity (Fg) is given by the equation Fg = mg, where m is the mass of the bob (725 g = 0.725 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The tension in the rope (T) can be split into the vertical component (Tv) and the horizontal component (Th). As the bob is in equilibrium, the vertical components of T and Fg must cancel each other out.
Using trigonometry, we can determine Tv = T * cos(θ), where θ is the angle the rope makes with the vertical. Therefore, we have:
Tv = Fg = mg
Substituting the known values, we get:
T * cos(33°) = (0.725 kg) * (9.8 m/s^2)
Now, solve for T using the equation above.
(b) To find the tension in the rope (T), we can use the vertical component of the tension force (Tv) calculated in part (a) and find the overall tension using Pythagoras' theorem. Since the bob is in equilibrium, the tension in the rope must balance the force of gravity.
Using the vertical component Tv and the horizontal component Th, we have:
Th = T * sin(θ)
Now, using Pythagoras' theorem, we can find the tension T:
T^2 = Tv^2 + Th^2
Solve for T using the equations above.