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.
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?
3 answers
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