your friend has a cheap attack and decides to hang his prized picture from as little string as possible, making the string almost horizontal. the picture is 50 lbs. because you have taken physics, you feel compelled to speak up. he says the string breaks at 75 lbs. so there’s no problem, but tell him the angle that he must have to avoid catastrophe…
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Assuming that there is an attachment point (into a wall) at the middle of string
{THIS should have been stated in question above :>}…then…
the weight (mg) of the picture must be supported by the two VERTICAL components (assumed equal in size) of the Tension in each half of the string. The CRITICAL angle, Θ, lies with its vertex at the attachment point between each half of string and the horizontal.
The static equilibrium equation in the VERTICAL direction is:
ΣFy = 0
50 = 2TsinΘ
and since Tmax = 75 this equation becomes:
50 = 2(75)(sin Θ)
sin Θ = 50/150
Θ = 19.5° <= critical angle
if Θ < 19.5°, sin Θ < 0.333 and T > 75 lbs ANS
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