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A uniform plank AB has mass 40 kg and length 4 m - Edexcel - A-Level Maths Mechanics - Question 1 - 2003 - Paper 1

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A uniform plank AB has mass 40 kg and length 4 m. It is supported in a horizontal position by two smooth pivots, one at the end A, the other at the point C of the pl... show full transcript

Worked Solution & Example Answer:A uniform plank AB has mass 40 kg and length 4 m - Edexcel - A-Level Maths Mechanics - Question 1 - 2003 - Paper 1

Step 1

the value of R

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Answer

To find the value of R, we first consider the forces acting on the plank. The total downward force acting on the system is the combined weight of the plank and the man:

extTotalDownwardForce=80g+40g=120g ext{Total Downward Force} = 80g + 40g = 120g

Using the equilibrium conditions, we can set the sum of forces in the vertical direction to zero:

RA+RB=120gR_A + R_B = 120g

Taking moments about point A (considering clockwise moments positive), we have:

  1. Moment due to the man (80 kg at 2 m from A): 80gimes280g imes 2
  2. Moment due to the plank's weight (40 kg at its center of gravity, which is 2 m from A): 40gimes240g imes 2

Setting the total moments equal to the moments around A:

Rimes4=80gimes2+40gimes2R imes 4 = 80g imes 2 + 40g imes 2 Rimes4=160g+80g=240gR imes 4 = 160g + 80g = 240g

Now, substituting for the total weight, we find:

R=240g4=60gR = \frac{240g}{4} = 60g

With gg typically approximated as 9.8extm/s29.8 ext{ m/s}^2, the value of R is:

R=60imes9.8=588extNR = 60 imes 9.8 = 588 ext{ N}

Step 2

the distance of the man from A

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Answer

To calculate the distance of the man from point A, we use the moment balance equation:

Setting the total clockwise moments around A equal to the total counterclockwise moments:

  1. Total weight's clockwise moments about A:

    80gimesx+40gimes2=60gimes380g imes x + 40g imes 2 = 60g imes 3

Now substituting gg into the equation simplifies to (using 60g60g as R):

80gimesx+80g=180g80g imes x + 80g = 180g

This simplifies to:

80gx=180g80g80gx = 180g - 80g 80gx=100g80gx = 100g

Dividing both sides by 80g80g gives:

x=10080extm=1.25extmx = \frac{100}{80} ext{ m} = 1.25 ext{ m}

Thus, the distance of the man from A is:

x=1.25extmx = 1.25 ext{ m}

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