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PHYSICS, M2 EQ-Bank 3

In the system diagram below, a 5-kilogram mass and masses \(A\) and \(B\) are held by high tensile frictionless wire in static equilibrium.
 

Using a vector diagram, calculate the masses of both \(A\) and \(B\).   (4 mark)

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\(\text{Mass}_A =3.91\ \text{kg}\)

\(\text{Mass}_B =4.55\ \text{kg}\)

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Using the sin rule both \(F_B\) and \(F_A\) can be calculated:

\(\dfrac{F_A}{\sin 48^{\circ}}\) \(=\dfrac{5 \times 9.8}{\sin 72^{\circ}}\)  
\(F_A\) \(=\dfrac{49\,\sin 48^{\circ}}{\sin 72^{\circ}}\)  
  \(=38.3\ \text{N}\)  

 
\(\text{Mass}_A =\dfrac{F}{a}=\dfrac{38.3}{9.8}=3.91\ \text{kg}\)

 

\(\dfrac{F_B}{\sin 60^{\circ}}\) \(=\dfrac{49}{\sin 72^{\circ}}\)
\(F_B\) \(=\dfrac{49\, \sin 60^{\circ}}{\sin 72^{\circ}}\)
  \(=44.6\ \text{N}\)

 

\(\text{Mass}_B =\dfrac{F}{a}=\dfrac{44.6}{9.8}=4.55\ \text{kg}\)

PHYSICS, M2 EQ-Bank 1

A concrete block with a weight of 1000 \(\text{N}\) is lifted with a pulley and chain system.

Initially the block is suspended by the chain as seen below. The block is then pulled to the side using a rope and makes an angle of 40° with the vertical as shown below. Ignore the masses of the chain and rope.
 

  1. State the tension in the chain initially.   (1 mark)
  2. Draw a free body diagram representing the forces acting on the concrete block in the final situation.   (2 marks)
  3. Determine the magnitude of the force being applied on the rope and the tension in the chain in the final situation.   (3 marks)
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a.    \(1000\ \text{N up.}\)

b. 
             

c.    \(F_{\text{rope}}=839\ \text{N}\)

       \(T_{\text{chain}}=1305.4\ \text{N}\)

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a.     → Tension is equal and opposite to the weight force of 1000 \(\text{N}\) down.

        → Hence the tension in the chain is 1000 \(\text{N}\) up.

 
b. 
             

 
c. 
       

Using the triangle above:

\(\tan 40^{\circ}\) \(=\dfrac{F_{\text{rope}}}{1000}\)  
\(F_{\text{rope}}\) \(=1000 \times \tan 40^{\circ}\)  
  \(=839\ \text{N}\)  

 

\( \cos 40^{\circ}\) \(=\dfrac{1000}{T_{\text{chain}}}\)  
\(T_{\text{chain}}\) \(=\dfrac{1000}{\cos 40^{\circ}}\)  
  \(=1305.4\ \text{N}\)