Moments (2) Flashcards
A non-uniform rod, AB, is freely hinged at a vertical wall. It is held horizontally in equilibrium by a strut attached to the wall at C at an angle of 55degrees. The thrust in the strut is 30 N as shown. The rod xm long and has a mass of 2kg, centred 0.4m from A. Take g = 9.8 ms^-2
Find the length of rod, x
Taking moments about A:
2g * 0.4 = 30cos55degrees * x
x = 7.84/17.20… = 0.456m
A non-uniform rod, AB, is freely hinged at a vertical wall. It is held horizontally in equilibrium by a strut attached to the wall at C at an angle of 55degrees. The thrust in the strut is 30 N as shown. The rod xm long and has a mass of 2kg, centred 0.4m from A. Take g = 9.8 ms^-2
Find the magnitude and direction of the reaction at A
Resolving vertically: R_V = 2g - 30cos55degrees = 2.392… N
Resolving horizontally: R_H = 30sin55degrees = 24.57… N
|R| = ¬(2.392…^2 + 24.57… ^2) = 24.7 N
tan pheta = 2.392… / 24.57… => pheta = 5.56degrees (3 s.f.)
A uniform rod of mass mkg rests in equilibrium against rough horizontal ground at Point A and a smooth peg at Point B, making an angle of pheta with the ground, where sin pheta = 3/5. The rod is lm long and B is (3/4)l from A
Show that the normal reaction at the peg, N = (2/3)mg cos pheta
Taking moments about A:
(1/2)l * mg cos*pheta = (3/4)l * N
N = ((1/2)l * mg cospheta) / (3/4)l = (2/3) * mg cospheta
A uniform rod of mass mkg rests in equilibrium against rough horizontal ground at Point A and a smooth peg at Point B, making an angle of pheta with the ground, where sin pheta = 3/5. The rod is lm long and B is (3/4)l from A
Find the range of possible values of the coefficient of friction between the rod and the ground
sin pheta = 3/5 => cos pheta = 4/5
Using the result of part a gives: N = (8/15)mg
Resolving horizontally:
F = N sin pheta = (8/15)mg * (3/5) = (8/25)mg
Resolving vertically:
R + N cos pheta = mg
R = mg - (4/5)N = mg - (32/75)mg = (43/75)mg
The rod is in non-limiting equilibrium so F <= μR:
So: (8/25)mg <= μ(43/75)mg => μ >= 24/43 = 0.56
A uniform ladder AB rests in limiting equilibrium against a smooth vertical wall (A) and upon the rough horizontal ground (B) at an angle of pheta. Clive stands on the ladder at point C a third of the way along its length from the base B
The ladder is 4.2m long and weighs 180 N. The normal reaction at A is 490 N. Given that tan pheta = 8/11.
Find the mass of Clive, m, to the nearest kg
Taking moments about the base of ladder B:
(mg cos pheta * 1.4) + (180 cos pheta * 2.1) = 490 sin pheta * 4.2
Dividing by cos pheta gives:
1.4mg + 378 = 2058 tan pheta = 2058 * 8/11
1.4mg = 1496.72… - 378 = 1118.72…
m = 1118.72… / 13.72 = 81.53… = 82kg
A uniform ladder AB rests in limiting equilibrium against a smooth vertical wall (A) and upon the rough horizontal ground (B) at an angle of pheta. Clive stands on the ladder at point C a third of the way along its length from the base B
The ladder is 4.2m long and weighs 180 N. The normal reaction at A is 490 N. Given that tan pheta = 8/11.
The coefficient of friction, μ, between the ground and the ladder
Resolving horizontally: F = 490 N
Resolving vertically: R = 180 + 81.53…g = 979.09… N
As equilibrium is limiting, F = μR
So: 979.09…μ = 490 N
μ = 0.50 (2 d.p.)
A uniform rod, AB, of mass 3kg is held horizontally in limiting equilibrium against a rough wall by an inextensible string connected to the road at point C and the wall at point D, as shown. A particle of mass mkg rests at point B. The magnitude of the normal reaction of the wall is 72.5 N.
Find the tension, T, in the string, and the mass, m, of the particle
tan pheta = 0.8/1.7
pheta = 25.20…degrees
Resolve horizontally:
T cos pheta = 72.5 N
t = 72.5 / cos25.20…degrees = 80.12… = 80.1 N
Taking moments about A:
(1.2 * 3g) + (2.4 * mg) = 1.7 * 80.12… sin25.20…degrees
23.52m = 58 - 35.28 = 22.72
m = 0.96598… = 0.966kg (3 s.f.)
A uniform rod, AB, of mass 3kg is held horizontally in limiting equilibrium against a rough wall by an inextensible string connected to the road at point C and the wall at point D, as shown. A particle of mass mkg rests at point B. The magnitude of the normal reaction of the wall is 72.5 N.
Find the frictional force, F, between the wall and the rod
Resolving vertically:
F + 80.12… sin25.20…degrees = 3g + 0.96598…g
F = 38.86… - 34.11… = 4.75 N (3 s.f.)
