State the principle of moments.
- (a)State the principle of moments.[2]
- (b)A solid plastic cylinder floats in water. It is used to support one end of a horizontal uniform beam AB as shown in [Figure 2.1]. The beam has length 6.0m and weight 1700N. The beam is attached to solid ground with a hinge at end A. The cylinder is floating vertically in the water. The top of the cylinder is attached at its centre to the beam at a horizontal distance of 5.0m from end A. The cylinder applies a vertical force of 1300 N to the beam. A person of weight 660 N stands on the beam at point P. The beam AB is in equilibrium.
- (b)(i)By taking moments about end A, determine the distance x from A to P. distance = ................................................................m [3][3]
- (b)(ii)The bottom of the cylinder is submerged in the water to depth y as shown in [Figure 2.2]. The beam is still attached to the cylinder but not shown. The cylinder has mass 11kg and diameter 0.78m. The beam exerts a vertical force of 1300N on the cylinder. The cylinder is in equilibrium. Show that the upthrust acting on the cylinder is 1400 N.[1]
- (b)(iii)The water has density 990kgm⁻³. Calculate the depth y. y = ................................................................m [2][2]
- (b)(iv)The person can stand anywhere between A and B. On [Figure 2.3], sketch the variation of the depth of the bottom of the cylinder with the distance of the person from A, for distances between 0 and 6.0 m. Numerical values are not required.[2]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.
- B1 in (rotational) equilibrium
- B1 sum/total of CW moments about a point = sum/total of ACW moments about the (same) point.
- C1 The magnitudes of the moments about A are: (1700 × 3.0) (660 x) (1300 × 5)
- C1 Correct magnitude of any one moment about A.
- A1 (1700 × 3.0) + (660 x) = (1300 × 5) x = 2.1 m
- A1 (Upthrust =) 11 × 9.81 + 1300 = 1400 (N)
- C1 y = Upthrust / pgA
- = 1400 / (990 × 9.81 × π× (0.78/2)²)
- A1 = 0.30 m
- OR
- [P = F / A = 1400 / π × (0.78 / 2)2 = 2930]
- y = Plpg
- = 2930 / (990 × 9.81)
- (C1)
- (A1) = 0.30 m
- B1 line with a non-zero value of depth at distance = 0
- B1 A line from distance = 0 to distance = 6.0 m with a gradient that is always positive
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