Question

ii.sketch on the diagram the average resultant force acting on the block between b and c. the arrow on the diagram represents the weight of the block. iii. calculate the magnitude of the average force exerted by the rope on the block between b and c.

Explanation:

Step1: Identify relevant forces

We need to consider the weight of the block and the force exerted by the rope to find the average resultant force. But without additional data like acceleration, mass etc., we assume basic force - balance concepts for the sketch. For the calculation in part iii, we assume we have mass \(m\) and acceleration \(a\) information later. Let the mass of the block be \(m\) and acceleration due to gravity \(g = 9.8\ m/s^{2}\), and assume an acceleration \(a\) of the block between B and C.

Step2: Sketch the average resultant force (for part ii)

The average resultant force \(\vec{F}_{net}\) acting on the block between B and C should be drawn starting from the center - of - mass of the block. If the block is accelerating upwards, the resultant force is in the upward direction. If it is accelerating downwards, the resultant force is in the downward direction. The direction depends on the motion of the block (e.g., if it is being lifted up, \(F_{net}=F_{rope}-mg\) and is upward; if it is falling with an acceleration \(a\), \(F_{net}=mg - F_{rope}\) and is downward).

Step3: Calculate the force exerted by the rope (for part iii)

Using Newton's second law \(F = ma\). Let the force exerted by the rope be \(F_{rope}\). If the block is moving with an acceleration \(a\) upwards, \(F_{rope}-mg=ma\), so \(F_{rope}=m(a + g)\). If the block is moving with an acceleration \(a\) downwards, \(mg - F_{rope}=ma\), so \(F_{rope}=m(g - a)\). But since no values of \(m\) and \(a\) are given in the problem statement, we leave the formula as \(F_{rope}=m(g\pm a)\) depending on the direction of acceleration.

Answer:

ii. Sketch an arrow starting from the center - of - mass of the block in the direction of the net acceleration (upward or downward depending on the motion of the block).
iii. \(F_{rope}=m(g\pm a)\) (plus sign if acceleration is upward, minus sign if acceleration is downward)