Question

In the system below, there are two masses labelled m and a 12 kg block that are connected to each other by two inextensible light strings B and C at point X. Point X is kept STATIONARY by attaching it to the wall using string A. The entire system is in EQUILIBRIUM. The coefficient of static friction between the 12 kg block and the surface of the table it is resting on is 0,53.

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Figure 5: Diagram forces in equilibrium. Question 3.1 [2 marks]
Define static friction.
Question 3.2 [3 marks]
Calculate the magnitude of the frictional force between the 12 kg block and the surface of the table.
Question 3.3 [5 marks]
Calculate the mass of body m.
Question 3.4 [1 mark]
Will the force in string A INCREASE, DECREASE or REMAIN THE SAME if string C was detached from the 12 kg block? State only the words INCREASE, DECREASE or REMAIN THE SAME.

Answer 1

[AI-generated answer — Bot GPT 3.5]

Question 3.1: Static friction is the force that prevents two surfaces from sliding past each other when they are in contact and at rest relative to each other.

Question 3.2: The magnitude of the frictional force between the 12 kg block and the surface can be calculated using the formula:
Frictional force = coefficient of static friction x Normal force
Frictional force = 0.53 x (12 kg x 9.81 m/s^2)
Frictional force = 61.78 N

Question 3.3: To calculate the mass of body m, we need to consider the forces acting on the system. Since the system is in equilibrium, the tension in string B is equal to the weight of the 12 kg block. Using this information, we can find the mass of body m to be:
m x 9.81 = 12 x 9.81
m = 12 kg

Question 3.4: INCREASE