Wednesday, September 27, 2017
q2 girl on pulley
2. (a) (pulley wheel) at rest / in equilibrium /
acceleration is zero B1
(b) (i) 500
N force down and general shape correct B1
angles
correct (one angle labelled correctly) B1
T1 and T2 directions labelled correctly B1
(ii) Formulae correct (resolving or sin rule) /
scale diagram drawn correctly
with
scale given B1
T1 = 674 (N) allow 650 to 700 for scale diagram A1
T2 = 766 (N) allow 740 to 790 for scale diagram A1
q1 masses on pulley
1.(a) (i) 1. mass = 360 / 9.8 36.7 (kg) B1
(allow 2sf)
(ii) 2. density
= mass / volume C1
= 36.7 / 4.7 ´ 10–3
= 7.8 ´ 10–3 A1
unit kg m–3 B1
(ii) right angled triangle with an additional
correct angle marked M1
set
of correct force labels and correct arrows A1
algebra
shown or scale given C1
tension
= 270 (N) or value in the range 255 to 285 (N) A1
(b) (i) tension is a vector / has magnitude and
direction B1
direction
involved in addition / the tensions or ropes act in
different directions B1
different directions B1
(ii) sum =270 sin37 + 360 sin53 B1
=162.5 + 287.5 B1
(or
one mark each for values of 162.5 and 287.5 seen) = 450 (N) A0
q3 Child on swing
3. (a) weight =
28 × 9.8 / mg C1
= 270 (N) (274.4) A1
(using g = 10 then –1)
= 270 (N) (274.4) A1
(using g = 10 then –1)
(b) a completed triangle drawn with correct
orientation B1
at least two labels for triangle with correct directions given B1
at least two labels for triangle with correct directions given B1
calculation: scale
diagram:
force P / weight = tan 35 scale given C1
force P =192 (N) 185 to 200 (N) A1
force P / weight = tan 35 scale given C1
force P =192 (N) 185 to 200 (N) A1
(c) tension
is greater B1
(reference to triangle) tension force would be greater (longer) as the holding
force P would be larger (longer) for greater angle / larger
value needed so vertical component still balances the weight B1
(reference to triangle) tension force would be greater (longer) as the holding
force P would be larger (longer) for greater angle / larger
value needed so vertical component still balances the weight B1
Monday, September 25, 2017
Resolving Forces
- A
weight of 50.0 N is suspended from a beam by a string. Calculate the
horizontal force must be applied to the weight to keep the string at an
angle of 40° to the vertical. 65.3N Calculate the tension in the
string.
- A
weight of 20 N rests on a plane inclined at 40° to the horizontal.
Calculate the components of the weight parallel to the plane and
perpendicular to the plane.12.86N,
15.32N
- A
weight of 4.33 N is suspended by a string
fastened at its upper end. A horizontal force is applied to the
weight so that the string makes an angle of 300
with the vertical. Calculate the force and the tension in
the string.2.50N, 5.00N
- A 10 N weight rests on a smooth
inclined plane. A force of 5N parallel to the plane
is needed to prevent the weight slipping down
the plane. Calculate the reaction of the
plane and the angle the plane makes with the horizontal.8.66N
,300
- A
garden roller of weight 800 N is being pulled along by a force of 200
N at 40° to the ground. Calculate (a) the force pulling the roller
forwards, 153.2N(b) the vertical force
exerted by the roller on the ground. 671.4 N
- Calculate
the forces (a) 153.2N and (b) 928.6N in the previous problem if the roller is
being pushed instead of being pulled.
Wednesday, September 13, 2017
Vectors worksheet
Solutions
1. 6.1N, 5.1N 2. -1.4N,
2.6N 3. 4.6N, -10.0N
4. -181.3N, -84.5N 5. -18.4N, -15.4N
Solutions
1. -6.1N, -5.1N 2. 1.1N,
-2.8N 3. -10.1N, 4.3N
4. 1.5N, 4.8N 5. 9.1N, 4.1N
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