Hint
Choose and retain a positive direction; distinguish signed quantities from magnitudes. Use the limiting equality and the horizontal normal reaction.
Worked solution
(a)
Use the limiting equality and the horizontal normal reaction.
At this limit \(\displaystyle P=F=\mu Mg=\)\(25\) N.
Answer: \(25\) N
(b)
Balance the unchanged vertical forces.
No vertical acceleration or angled pull gives \(\displaystyle R=Mg=50 N\).
Answer: \(50\) N
(c)
Subtract sliding friction and divide the resultant by mass.
The resultant is \(45\)−\(25\)=20 N, so \(\displaystyle a= \frac{4m}{s^{2}}\).
Answer: \(4\) \(\displaystyle \frac{m}{s^{2}}\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternative methods and consequential reasoning. An unexplained result does not earn a method mark.
(a) · 2 marks
- 1 mark: Use the limiting equality and the horizontal normal reaction.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Equality \(\displaystyle F=\mu R\) is appropriate at the stated limit, not at every equilibrium.
(b) · 2 marks
- 1 mark: Balance the unchanged vertical forces.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: The driving pull is not the resultant once sliding friction is present.
(c) · 2 marks
- 1 mark: Subtract sliding friction and divide the resultant by mass.
- 1 mark: Give the correct requested result, retaining signs, units and any domain restrictions.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Equality \(\displaystyle F=\mu R\) is appropriate at the stated limit, not at every equilibrium.
Total: 6 marks.