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A steel girder AB, of mass 200 kg and length 12 m, rests horizontally in equilibrium on two smooth supports at C and D, where AC = 2 m and DB = 2 m - Edexcel - A-Level Maths Mechanics - Question 2 - 2013 - Paper 1
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A steel girder AB, of mass 200 kg and length 12 m, rests horizontally in equilibrium on two smooth supports at C and D, where AC = 2 m and DB = 2 m. A man of mass 80... show full transcript
Worked Solution & Example Answer:A steel girder AB, of mass 200 kg and length 12 m, rests horizontally in equilibrium on two smooth supports at C and D, where AC = 2 m and DB = 2 m - Edexcel - A-Level Maths Mechanics - Question 2 - 2013 - Paper 1
Step 1
(a) Find the magnitude of the reaction on the girder at the support at C.
Answer
To find the reaction at support C, we need to analyze the moments about one support. We will sum the moments around point C.
Let ( R_C ) be the reaction at support C.
The total weight acting downwards includes the weight of the girder and the man:
- Weight of the girder: ( 200 \text{ kg} \times 9.81 = 1962 \text{ N} )
- Weight of the man: ( 80 \text{ kg} \times 9.81 = 784 \text{ N} )
Total weight ( = 1962 + 784 = 2746 \text{ N} )
The sum of moments about point C equals zero because the system is in equilibrium:
[ R_C \times 10 = 784 \times 6 + 200 \times 4 ] [ 10 R_C = 4704 + 800 ] [ 10 R_C = 5504 ] [ R_C = 550.4 \text{ N} ]
Thus, the reaction at the support at C is approximately ( R_C \approx 550.4 \text{ N} ).
Step 2
(b) the magnitude of the reaction at the support at X
Answer
Since the reactions at the two supports are equal and the girder is in equilibrium:
Let ( R_X ) be the reaction at support X. As found earlier, we will set up the equilibrium conditions again. The sum of upward forces must equal the sum of downward forces:
[ R_C = R_X ]
Using the previously derived total weight, we know:
[ 2R_X = 2746 \text{ N} ] [ R_X = \frac{2746}{2} = 1373 \text{ N} ]
Thus, the reaction at the support at X is ( R_X \approx 1373 \text{ N} ).
Step 3
(c) the value of x.
Answer
Now, we need to find the value of x. Again, using the moments about support C, we will write:
At equilibrium, we can set the moments about point B:
[ S_x + (S_X \times 10) = (80 \times 8) + (200 \times 6) ]
Where S is the reaction at C and S_X is the reaction at X:
[ 140 + 1400 = 640 + 1200 ] [ 140 + 1400 = 1840 ] [ 140 = 440 \text{ Field equals} \frac{x}{12} ]
Therefore, solving yields ( x = \frac{12}{2} = 6 \text{ m} ).