A 100kg, 2m long plank sticks horizontally out of a wall. A 600N force is applied at the other end at the following angles:
a) 60o
b) 120o
c) 0o
Find the NET TORQUE about the attachment point for each. SHOW ALL WORK.
Solution)
Given,
Mass, m=100 kg
Length, r=2 m
Force, F=600 N
Part a)
We know, Torque, T= r*Fsintheta
Substitute values,
T=2*600*sin60= 1039.23 Nm (Ans)
========
Part b)
Theta=120°
T=2*600*sin120= 1039.23 Nm (Ans)
========
Part c)
Thet=0°
So, Torque=0 (since sin0=0)
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A 100kg, 2m long plank sticks horizontally out of a wall. A 600N force is applied...
A 100-kg, 2-m long plank sticks horizontally out of a wall. A 600-N force is applied at the other end at the following angles: a) 60° b) 120° c) 0° FIND THE NET TORQUE ABOUT THE ATTACHMENT POINT FOR EACH. SHOW ALL WORK
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