Question
Fig. 2.1 shows a sign that extends over a road.
The mass of the sign is \(3.4 × 10^3\) kg.
(a) Calculate the weight W of the sign.
W =
(b) The weight of the sign acts at a horizontal distance of 1.8 m from the centre of the support post and it produces a turning effect about point P.
Point P is a horizontal distance of 1.3 m from the centre of the support post.
(i) Calculate the moment about P due to the weight of the sign.
moment =
(ii) A concrete block is positioned on the other side of the support post with its centre of mass a horizontal distance of 70 cm from the centre of the support post.
1. State what is meant by centre of mass.
2. The weight of the concrete block produces a moment about point P that exactly cancels the moment caused by the weight W.
Calculate the weight of the concrete block.
weight =
(c) The concrete block is removed. The sign and support post rotate about point P in a clockwise direction.
State and explain what happens to the moment about point P due to the weight of the sign as it rotates.
Answer/Explanation
Answer:
(a) (W =) mg OR \(3.4 × 10^3 × 10\)
\(3.4 × 10^4\) N
(b) (i) moment = Fx in any form OR (moment) = Fx OR 0.50 (seen)
\(3.4 × 10^4 × (1.8 – 1.3)\) OR \(3.4 × 10^4 × 0.50\)
\(1.7 × 10^4\) N m
(ii) 1. (the point) where (all) the mass can be considered to be concentrated
2. \(1.7 × 10^4 / (1.3 + 0.70)\) OR \(1.7 × 10^4 / (2.0)\)
\(8.5 × 10^3\) N
(c) (moment / it) increases perpendicular distance (between P and line of action of) W increases