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PHYSICS, M2 EQ-Bank 10

A 20 kg crate is suspended in equilibrium by two cables as seen in the diagram below:
 

 

Determine the tension force \((T_2)\) in Cable 2.   (3 marks)

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\(17.2\ \text{N}\)

Show Worked Solution
  • The weight force of the crate, \(F_w=20 \times 9.8 = 19.6\ \text{N}\) and so the following vector diagram can be set up below.

  • Using the \(\sin\) rule for finding the side length of a triangle, \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\)
\(\dfrac{T_2}{\sin 60}\) \(=\dfrac{F_w}{\sin 80}\)  
\(T_2\) \(=\dfrac{19.6}{\sin 80} \times \sin 60\)  
  \(=17.2\ \text{N}\)  

PHYSICS, M2 EQ-Bank 9

The diagram below shows a 20 kg mass suspended from two strings.
 
 
 
Using a force diagram, calculate the force \(F\) in each string acting on the mass.   (3 marks)
  
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\(196\ \text{N}\)

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  • The weight force of the mass  \(F_w = mg = 20 \times 9.8 = 196\ \text{N}\).
  • As the force diagram is an equilateral triangle, the side length of all the sides of the triangle are equal.
  • Hence the magnitude of \(F_w = F = 196\ \text{N}\).
  • The force in each string is \(196\ \text{N}\).

PHYSICS, M2 EQ-Bank 7

A 500 gram block is placed on a frictionless inclined plane and joined to a 250 gram block by a tensile frictionless wire, as shown in the diagram below.
 

Determine the angle of inclination at which the system remains in static equilibrium.   (2 marks)

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\(30^{\circ}\)

Show Worked Solution
  • For the system to be in static equilibrium, the force of the 500 g block, \((m_1)\), down the slope must be equal to the weight force of the 250 g block, \((m_2)\).
\(m_1g \times \sin \theta\) \(=m_2g\)  
\(\sin \theta\) \(=\dfrac{m_2}{m_1}\)  
\(\theta\) \(=\sin^{-1}\left(\dfrac{0.250}{0.5}\right) = 30^{\circ}\)  

 

  • The angle of inclination needs to be \(30^{\circ}\).

PHYSICS, M2 EQ-Bank 6

A worker is using a pulley system to lift and lower a container at a construction site. The empty container has a mass of 5.0 kg, and when loaded with bricks, its mass increases to 18 kg.

The container is lowered vertically with a constant acceleration of 3.2 m/s\(^2\). Later, it is lifted back up at a constant speed of 2.8 m/s.

  1. Calculate the tension in the rope while the container is being lowered.   (2 marks)
  1. Calculate the tension in the rope while the container is being raised.   (1 mark)
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a.    \(33\ \text{N}\)

b.    \(147\ \text{N}\)

Show Worked Solution

a.   Tension while container lowered:

  • Force due to gravity acts in the opposite direction to the tension force in the rope.
  • Total force on the system \(= F_g-T\)
\(ma\) \(=mg-T\)  
\(T\) \(=mg-ma\)  
  \(= 5 \times 9.8-5 \times 3.2\)  
  \(= 33\ \text{N}\)  

 
b.   Tension while container raised:

Total force = \(0\ \text{N}\)  (raised at a constant speed)

\(mg-T\) \(=0\)  
\(\therefore T\) \(=mg = 15 \times 9.8 = 147\ \text{N}\)  

PHYSICS, M2 EQ-Bank 5

Consider the system below:
 

Two blocks are connected by a light, inextensible string that passes over a frictionless pulley, as shown in the diagram.

The 3.0 kg block rests on a smooth, horizontal surface, while the other 6.0 kg block hangs vertically off the edge of the table.

  1.  Calculate the acceleration of the entire system.   (3 marks)
  1. Hence, determine the magnitude of the tension, \(T\), in the string at point \(X\).   (1 mark)
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a.    \(6.53\ \text{ms}^{-2}\)

b.    \(19.6\ \text{N}\)

Show Worked Solution

a.   \(F_{\text{net}} = 3\ \text{kg} \times a_{\text{system}} = T\)

\(F_{\text{net}} = 6\ \text{kg} \times a_{\text{system}} = 6\ \text{kg} \times 9.8-T\)   \((F_w-T)\)
 

Substitute the first equation into the second equation:

\(6\ \text{kg} \times a_{\text{system}}\) \(=6\ \text{kg} \times 9.8-(3\ \text{kg} \times a_{\text{system}})\)  
\(6\ \text{kg} \times a_{\text{system}} + 3\ \text{kg} \times a_{\text{system}}\) \(=58.8\ \text{N}\)  
\(a_{\text{system}} \times (6+3)\ \text{kg}\) \(=58.8\ \text{N}\)  
\(a_{\text{system}}\) \(=\dfrac{58.8\ \text{N}}{9\ \text{kg}}\)  
  \(=6.53\ \text{ms}^{-2}\)  

 

b.    \(T= 3\ \text{kg} \times a_{\text{system}} = 3 \times 6.53 = 19.59 = 19.6\ \text{N}\)

PHYSICS, M2 EQ-Bank 2 MC

A 25 kg object is hanging from a string and is being pulled sideways by a force \((F_1)\), so that the string makes an angle of 40° with the vertical, as shown in the diagram.

Assuming the object remains stationary, determine the magnitude of the force \((F_1)\) pulling the object?
 

  1. 115 N
  2. 157 N
  3. 191 N
  4. 206 N
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\(D\)

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\(\therefore \tan 40^{\circ}\) \(=\dfrac{F_1}{F_w}\)  
\(F_1\) \(=F_w \times \tan 40^{\circ}\)  
  \(= 25 \times 9.8 \times \tan 40^{\circ}\)  
  \(= 205.6\ \text{N}\)  

 
\(\Rightarrow D\)

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

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

Show Worked Solution

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.   \(F_{\text{rope}}=839\ \text{N}\)

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

Show Worked Solution

a.   Initial tension in chain:

  • 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. 
       

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

 

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