Question

Show work and answer please.

Part A the cord makes a4 angle with A heavy steel ball is hung from a cord to make a pendulum. The ball is pulled to the side so that the vertical., Holding the ball in place takes a force of 20 N. If the ball is pulled farther to the side so that the cord makes a 6 angle, what force is required to hold the bal? Express your answer with the appropriate units IA F- Value Units Submit Request Answer

Answer 1

Consider Case 1(4 deg) Consider all horizontal forces sigma F-x = 0 T_1 sin 4 = 20 N T_1 286.71 N Consider all vertices forces sigma F_y = 0 T_1 cos 4 = mg 286.71 N * cos 4 = mg mg = 286 N So Weight of steel balls = 286 N Consider Case 2(6 deg) Consider all vertical forces sigma F_y = 0 T_2 cos 6 = mg T_2 = mg/cos 6 T_2 = 286 N/cos 6 T_2 = 287.58 N Consider all horizontal forces sigma F_x = 0 T_2 sin 6 = F 287.58 N * sin 6 = F ANSWER: F = 30.06 N

Consider Case 1(4 deg)

Consider all horizontal forces

$\sum F_{x}=0$

$T_{1}sin4=20N$

$T_{1}=286.71N$

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Consider all vertical forces

$\sum F_{y}=0$

$T_{1}cos4=mg$

$286.71N*cos4=mg$

$mg=286N$

so Weight of steel ball =286N

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Consider Case 2(6 deg)

Consider all vertical forces

$\sum F_{y}=0$

$T_{2}cos6=mg$

$T_{2}=mg/cos6$

$T_{2}=286N/cos6$

$T_{2}=287.58N$

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Consider all horizontal forces

$\sum F_{x}=0$

$T_{2}sin6=F$

$287.58N*sin6=F$

ANSWER: $F=30.06N$

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