Question 19M.3.SL.TZ2.8
| Date | May 2019 | Marks available | [Maximum mark: 5] | Reference code | 19M.3.SL.TZ2.8 |
| Level | SL | Paper | 3 | Time zone | TZ2 |
| Command term | Calculate, Determine, Outline | Question number | 8 | Adapted from | N/A |
A uniform ladder of weight 50.0 N and length 4.00 m is placed against a frictionless wall making an angle of 60.0° with the ground.
Outline why the normal force acting on the ladder at the point of contact with the wall is equal to the frictional force F between the ladder and the ground.
[1]
«translational equilibrium demands that the» resultant force in the horizontal direction must be zero✔
«hence NW = F»
Equality of forces is given, look for reason why.
Many candidates stated that the resultant of all forces must be zero but failed to mention the fact that horizontal forces must balance in this particular question.
Calculate F.
[2]
«clockwise moments = anticlockwise moments»
50 × 2cos 60 = NW × 4sin 60 ✔
«»
F = 14.4«N» ✔
Very few candidates could take moments about any point and correct answers were rare both at SL and HL.
The coefficient of friction between the ladder and the ground is 0.400. Determine whether the ladder will slip.
[2]
maximum friction force = «0.4 × 50N» = 20«N» ✔
14.4 < 20 AND so will not slip ✔
The question about the slipping of the ladder was poorly answered. The fact that the normal reaction on the floor was 50N was not known to many.