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A 2.15-kg object hangs in equilibrium at the end of a rope (taken as massless) while...

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.

Solutions

Expert Solution

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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genius_generous answered 7 months ago
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