A beam AB has mass 12 kg and length 5 m. It is held in equilibrium in a horizontal position by two vertical ropes attached to the beam. One rope is attached to A, the other to the point C on the beam, where BC = 1 m, as shown in Figure 2. The beam is modelled as a uniform rod, and the ropes as light strings. a) Find i) the tension in the rope at C; ii) the tension in the rope at A. A small load of mass 16 kg is attached to the beam at a point which is y metres from A. The load is modelled as a particle. Given that the beam remains in equilibrium in a horizontal position, b) find, in terms of y, an expression for the tension in the rope at C. c) The rope at C will break if its tension exceeds 98 N. The rope at A cannot break. Find the range of possible positions on the beam where the load can be attached without the rope at C breaking.

Photo AI

A beam AB has mass 12 kg and length 5 m - Edexcel - A-Level Maths Mechanics - Question 5 - 2008 - Paper 1

Question 5

A beam AB has mass 12 kg and length 5 m. It is held in equilibrium in a horizontal position by two vertical ropes attached to the beam. One rope is attached to A, th... show full transcript

Worked Solution & Example Answer:A beam AB has mass 12 kg and length 5 m - Edexcel - A-Level Maths Mechanics - Question 5 - 2008 - Paper 1

Step 1

Find (i) the tension in the rope at C;

96%

114 rated

Answer

To find the tension at point C, we first need to consider the forces acting on the beam. The weight of the beam acts at its midpoint, which is at 2.5 m from point A. The total weight (W) can be calculated as:

$W = m imes g = 12 ext{ kg} imes 9.8 ext{ m/s}^2 = 117.6 ext{ N}$

Taking moments about point A:

$T imes 4 = W imes 2.5$

Substituting the values:

$T imes 4 = 117.6 ext{ N} imes 2.5$

Solving for T gives:

$T = \frac{117.6 ext{ N} imes 2.5}{4} = 73.5 ext{ N}$

Step 2

Find (ii) the tension in the rope at A;

99%

104 rated

Answer

Next, to calculate the tension at point A, we can use the equilibrium of forces. The total upward forces must equal the total downward forces:

$S + T = 12g$

Where S is the tension at A. Substituting in T:

$S + 73.5 = 12 imes 9.8$

Calculating this gives:

$S + 73.5 = 117.6$

Thus,

$S = 117.6 - 73.5 = 44.1 ext{ N}$

Step 3

Find in terms of y, an expression for the tension in the rope at C.

96%

101 rated

Answer

When a load of mass 16 kg is added at a distance y from A, the additional weight (W_load) is given by:

$W_{load} = 16 ext{ kg} imes 9.8 ext{ m/s}^2 = 156.8 ext{ N}$

The total moment about point A becomes:

$T imes 4 = (117.6 + 156.8) imes y$

So, we can write:

$T = \frac{(117.6 + 156.8) imes y}{4} = \frac{274.4y}{4} = 68.6y$

Step 4

Find the range of possible positions on the beam where the load can be attached without the rope at C breaking.

98%

120 rated

Answer

The rope at C will break if:

$T > 98 ext{ N}$

Setting our expression for T gives:

$68.6y > 98$

Therefore,

$y > \frac{98}{68.6} \\ y > 1.43 ext{ m}$

Since y must be no more than 5 m from A, the load must be placed more than 1.43 m but less than or equal to 5 m,

The range of possible positions:

$1.43 < y \leq 5 ext{ m}$

A beam AB has mass 12 kg and length 5 m - Edexcel - A-Level Maths Mechanics - Question 5 - 2008 - Paper 1

Worked Solution & Example Answer:A beam AB has mass 12 kg and length 5 m - Edexcel - A-Level Maths Mechanics - Question 5 - 2008 - Paper 1

Find (i) the tension in the rope at C;

Find (ii) the tension in the rope at A;

Find in terms of y, an expression for the tension in the rope at C.

Find the range of possible positions on the beam where the load can be attached without the rope at C breaking.

Join the A-Level students using SimpleStudy...