In: Physics
A 2.15-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 18.5-N horizontal force. Find the magnitude of the tension in the rope, and the rope\'s angle from the vertical. The acceleration due to gravity is 9.81 m/s2.
In the FBD above,
F is the force of the wind,
T is the tension in the rope,
mg is the weight of the mass,
is the angle,
that the rope makes with the horizontal
doing horizontal force balance
F = T*sin()------(1)
and from vertical force balance
mg = T*cos()------(2)
squaring and adding both the equations,
T2 = F2 + (mg)2
T = sqrt(18.52 + (2.15*9.81)2)
T = 20.94 N
Now, dividing equation 1 and 2
tan() = F/mg =
18.5/(2.15*9.81)
=
tan-1(18.5/(2.15*9.81))
=
41.255o
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