1. The problem statement, all variables and given/known data You are holding a sign as shown below. The sign (including the horizontal bar it hangs from) has a mass of 2.20 kg and is 40.5 cm wide. The sign is hanging from a 1.03 m tall, 4.46 kg vertical post. The sign is symmetric with a uniform mass distribution as implied by the drawing What external force (+ if up, - if down) and torque (+ for CCW, - for CW) do you have to apply with your hand to keep the sign in static equilibrium? [Use g = 9.80 m/s2 and ignore the width of the vertical post.] 2. Relevant equations Net Forces = 0 Torques (counter clockwise) = Torques (clockwise) 3. The attempt at a solution I think that the only forces acting on the sign are vertical so for the net forces equation I have: Forces up = Forces down =====> FHand = Mg + mg Where M is the mass of the vertical post and m is the mass of the sign. For the conservation of torques I have: d * FHand = d*Mg + d*mg Where d is the lever arm in meters. (The axis of rotation is at the left side.) 0.405 * FHand = 0.405 * (4.46kg*9.8) + 0.2025 * (2.20kg*9.8) FHand = (17.7017 N*m + 4.3659 N*m) / 0.405 m FHand = 54.488 N Torque = 22.0676 N*m This answer is wrong and I do not know why.