FIll in the blanks! d = 4.7 cm These two structures are similar to what you buil
ID: 3278224 • Letter: F
Question
FIll in the blanks! d = 4.7 cm
These two structures are similar to what you built except we have cords instead of springs. Use the weight of your bar for "w" and for "F", use the force you had in the lab by hanging the 350[g] mass on the bar. In figure A it's possible to solve for "d" because you'll have three unknowns (T_1, T_2, d) and three equations (Sigma tau = 0 and Newton's 2^nd Law for x and y components). So solve that: T_1 = _______ T_2 = _______ d = _______ In figure 2 it won't be so easy since Newton's 2^nd Law for the x-components will be 0=0 no matter what the tension forces are. In fact you obviously could change the value of "d" and since the cords don't stretch, the bar would still hang straight. Knowing "d" you could solve for T_1 and T_2 If d=0.060[m] find the tension forces: T_1 = ________ T_2 = _______Explanation / Answer
From the figure and the given data
weight of the hanging bar, w = 3.47 N
Applied force, F = 67.4 N
let tension in the left stirng be T1 and in the right be T2
for figure 1
From force balance
T1cos(30) + T2*cos(20) = w + F
T1sin(30) = T2sin(20)
from moment balance about left end of the rod
w*0.09 + F(0.09+d) = T2*cos(20)*0.17
3.47*0.09 + 67.4(0.09 + d) = T2*0.159
but T1 = 0.684T2
so, 0.684T2*cos(30) + T2*cos(20) = 3.47 + 67.4
T2 = 46.258 N
T1 = 31.64 N
and d = 0.0144 m
from second diagram
so from force balance
T1 + T2 = w + F
also, from moment balance about left end
w*0.09 + f*(0.09 + d) = T2*0.17
now, d= 0.06 m
T2 = 61.307 N
T1 = 9.562 N
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