3.1 Define tension force, and give an example of such a force in the diagram above - NSC Technical Sciences - Question 3 - 2020 - Paper 1

DECREASES

The free-body diagram should include the following forces:

1. Weight (Fg) acting downwards.
2. Normal force (FN) acting upwards.
3. Frictional force (fk) acting opposite to the direction of motion.
4. Force from Tom (FTom) acting horizontally to the right (200 N).
5. Force from Zane (FZane) acting at an angle of 65° with the horizontal, which can be broken into horizontal (FZane Cos 65°) and vertical components (FZane Sin 65°).

To find the coefficient of kinetic friction, we can use the following formula:

$F_{net} = F_{Tom} + F_{Zane ext{Cos }65°} - f_{k}$

Given that the net force (F_net) is 205 N, we have:

$205 = 200 + 160 imes ext{Cos }65° - f_{k}$

Calculating the forces:

1. Calculate FZane Cos 65°: $$$$

ightarrow 67.6336 N$$

2. Substitute to find fk: $205 = 200 + 67.6336 - fk$ $fk = 267.6336 - 205$ $fk = 62.6336 N$

3. Now, we find the normal force (FN): $FN = mg - F_{Zane} ext{Sin }65°$ (where FZane Sin 65° acts vertically)

   $$$$

ightarrow 147.668 N$$

Thus, the normal force is: $FN = 350 - 147.668 = 202.332 N$

4. Finally, calculate μk: $$$$

ightarrow ext{μk} = rac{62.6336}{202.332} ightarrow 0.3094$$

Therefore, the coefficient of kinetic friction is approximately 0.3094.