Moments Flashcards
A steel girder AB, of mass 200 kg and length 12 m, rests horizontally in equilibrium on two smooth supports at C and at D, where AC = 2 m and DB = 2 m. A man of mass 80 kg stands on the girder at the point P, where AP = 4 m, as shown in Figure 1.
( Figure shows a " rod " AB which is 12m ) ( C and D are supports ) ( C is 2m from A ) ( D is 2m from B ) ( The child " P " is 4m from A ) ( P has a weight of 80kg ) ( Rod has a weight of 200kg ) ( Uniform rod, so 200kg is 6m away from both A and B )
The man is modelled as a particle and the girder is modelled as a uniform rod.
Find the magnitude of the reaction on the girder at the support at C.
D: ( Taking moments from D )
- ( 4 x 200g ) + ( 6 x 80g ) = ( 8 x R )
- ( Downwards force = upwards force )
- 8R = 1260g
- R = 160g
The support at D is now moved to the point X on the girder, where XB = x metres.
The man remains on the girder at P, as shown in Figure 2.
( Figure shows a " rod " AB which is 12m ) ( C and X are supports ) ( C is 2m from A ) ( X is xm from B ) ( The child " P " is 4m from A ) ( P has a weight of 80kg ) ( Rod has a weight of 200kg ) ( Uniform rod, so 200kg is 6m away from both A and B )
Given that the magnitudes of the reactions at the two supports are now equal and that the girder again rests horizontally in equilibrium, find the magnitude of the reaction at the support at X.
- 2R = 200g + 80g
- R = 140g
The support at D is now moved to the point X on the girder, where XB = x metres.
The man remains on the girder at P, as shown in Figure 2.
( Figure shows a " rod " AB which is 12m ) ( C and X are supports ) ( C is 2m from A ) ( X is xm from B ) ( The child " P " is 4m from A ) ( P has a weight of 80kg ) ( Rod has a weight of 200kg ) ( Uniform rod, so 200kg is 6m away from both A and B )
Given that the magnitudes of the reactions at the two supports are now equal and that the girder again rests horizontally in equilibrium, find the value of x.
( X = 140g )
B: ( Taking moments from B )
- ( x x 140g ) + ( 10 x 140g ) = ( 6 x 200g ) + ( 8 x 80g )
- 140gx + 1400g = 1200g + 640g
- 140gx = 440g
- 140x = 440
- x = 3.142857143
- x = 3.14m
State two ways in which you have used the assumptions made in modelling the plank as a uniform rod.
- Weight of the plank is at the centre
- Plank doesn’t bend
A beam AB has weight W newtons and length 4 m.
The beam is held in equilibrium in a horizontal position by two vertical ropes attached to the beam.
One rope is attached to A and the other rope is attached to the point C on the beam, where AC = d metres, as shown in
Figure 4.
The beam is modelled as a uniform rod and the ropes as light inextensible strings.
The tension in the rope attached at C is double the tension in the rope attached at A.
( Figure 4 shows a beam AB ) ( AB is 4m ) ( A and C are both ropes ) ( Beam has a weight of W ) ( Uniform rod, so W is 2m form A and B ) ( AC is dm )
Find the value of d.
- 2R + R = W
- 3R = W
- R = W / 3
A:
- ( 2 x W ) = ( ( x + 2 ) x 2R )
- ( x is the distance from the midpoint to C )
- 2W = 2Rx + 4R
- 2W = 2( W / 3 )x + 4( W / 3 )
- 2W = 2Wx / 3 + 4W / 3
- 2W / 3 = 2Wx / 3
- x = 1
- d = 2 + 1
- d = 3m
A small load of weight kW newtons is attached to the beam at B. The beam remains in equilibrium in a horizontal position.
The load is modelled as a particle.
The tension in the rope attached at C is now four times the tension in the rope attached at A.
( Figure 4 shows a beam AB ) ( AB is 4m ) ( A and C are both ropes ) ( Beam has a weight of W ) ( Uniform rod, so W is 2m form A and B ) ( AC is 3m ) ( Weight kW is at B )
Find the value of k.
- 4R + R = W + kW
- 5R = W + kW
- R = W + kW / 5
A:
- ( 2 x W ) + ( 4 x kW ) = ( 3 x 4R )
- 2W + 4kW = 12R
- 2W + 4kW = 12( W + kW / 5 )
- 2W + 4kW = 12W + 12kW / 5
- 10W + 20kW = 12W + 12kW
- 8kW = 2W
- 8k = 2
- k = 1 / 4
A plank PQR, of length 8 m and mass 20 kg, is in equilibrium in a horizontal position on two supports at P and Q, where PQ = 6 m.
A child of mass 40 kg stands on the plank at a distance of 2 m from P and a block of mass Mkg is placed on the plank at the end R.
The plank remains horizontal and in equilibrium.
The force exerted on the plank by the support at P is equal to the force exerted on the plank by the support at Q.
By modelling the plank as a uniform rod, and the child and the block as particles, find the magnitude of the force exerted on the plank by the support at P.
- ( Drawn diagram of a rod PR )
- ( P and Q are both supports )
- ( A child with a mass of 40kg is 2m from P )
- ( PQ is 6m apart )
- ( Weight of Mkg is at R )
- ( PR is 8m )
- ( Rod has a mass of 20kg )
- ( Both supports have a force of x )
- ( Drawn diagram of a rod PR )
- ( P and Q are both supports )
- ( A child with a mass of 40kg is 2m from P )
- ( PQ is 6m apart )
- ( Weight of Mkg is at R )
- ( PR is 8m )
- ( Rod has a mass of 20kg )
- ( Both supports have a force of x )
R:
- ( 4 x 20g ) + ( 6 x 40g ) = ( 8 x x ) + ( 2 x x )
- 10x = 80g + 240g
- 10x = 320g
- x = 32g
A plank PQR, of length 8 m and mass 20 kg, is in equilibrium in a horizontal position on two supports at P and Q, where PQ = 6 m.
A child of mass 40 kg stands on the plank at a distance of 2 m from P and a block of mass Mkg is placed on the plank at the end R.
The plank remains horizontal and in equilibrium.
The force exerted on the plank by the support at P is equal to the force exerted on the plank by the support at Q.
By modelling the plank as a uniform rod, and the child and the block as particles, find the value of M.
- ( Drawn diagram of a rod PR )
- ( P and Q are both supports )
- ( A child with a mass of 40kg is 2m from P )
- ( PQ is 6m apart )
- ( Weight of Mkg is at R )
- ( PR is 8m )
- ( Rod has a mass of 20kg )
- ( Both supports have a force of x )
- ( x = 32g )
- ( Drawn diagram of a rod PR )
- ( P and Q are both supports )
- ( A child with a mass of 40kg is 2m from P )
- ( PQ is 6m apart )
- ( Weight of Mkg is at R )
- ( PR is 8m )
- ( Rod has a mass of 20kg )
- ( Both supports have a force of x )
- ( x = 32g )
- 2x = 40g + 20g + Mg
- ( Upwards force = downwards force )
- 2( 32g ) = 60g + Mg
- 64g = 60g + Mg
- Mg = 4g
- M = 4
State how, in your calculations, you have used the fact that the child and the block can be modelled as particles.
- Weights act at a point
A non-uniform plank AB has length 6 m and mass 30 kg.
The plank rests in equilibrium in a horizontal position on supports at the points S and T of the plank where AS = 0.5m and TB = 2 m.
When a block of mass M kg is placed on the plank at A, the plank remains horizontal and in equilibrium and the plank is on the point of tilting about S.
When the block is moved to B, the plank remains horizontal and in equilibrium and the plank is on the point of tilting about T.
The distance of the centre of mass of the plank from A is d metres. The block is modelled as a particle and the plank is modelled as a non-uniform rod.
Find the value of d.
( Drawn diagram of a rod AB ) ( AB is 6m apart ) ( Supports S and T are present ) ( Mkg is labelled 1 at A ) ( Mkg is labelled 2 at B ) ( Rod has a weight of 30kg ) ( S is 0.5m from A ) ( A to the weight of the rod is dm ) ( T is 2m from B )
S: ( When tilting on S )
- ( ( d - 0.5 ) x 30g ) = ( 0.5 x Mg )
- Mg / 2 = 30gd - 15g
- Mg = 60gd - 30g
T: ( When tilting on T )
- ( 2 x Mg ) = ( ( 4 - d ) x 30g )
- 2mg = 120g - 30gd
- Mg = 60g - 15gd
- ( Form a simultaneous equation )
- Mg = 60gd - 30g
- Mg = - 15gd + 60g
- 0 = 75gd - 90g
- 90g = 75gd
- d = 1.2m
A non-uniform beam AD has weight W newtons and length 4 m. It is held in equilibrium in a horizontal position by two vertical ropes attached to the beam.
The ropes are attached to two points B and C on the beam, where AB = 1 m and CD = 1 m, as shown in Figure 7.
The tension in the rope attached to C is double the tension in the rope attached to B.
The beam is modelled as a rod and the ropes are modelled as light inextensible strings.
A small load of weight kW newtons is attached to the beam at D. The beam remains in equilibrium in a horizontal position.
The load is modelled as a particle.
Find an expression for the tension in the rope attached to B, giving your answer in terms of k and W.
( Figure shows a rod AD ) ( AD is 4m apart ) ( Ropes are attached to B and C ) ( B is 1m from A ) ( C is 1m from B ) ( Weight of beam is WN ) ( D has a weight of kWN ) ( BW is 4 / 3m ) ( CW is 2 / 3m ) ( Ropes have a force of R and 2R for B and C ) ( BC is 2m apart )
C:
- ( TB x 2 ) + ( kW x 1 ) = ( W x 2 / 3 )
- 2TB + Kw = 2W / 3
- 2TB = 2W / 3 - kW
- 2TB = 2W - 3kW / 3
- TB = 2W - 3kW / 6
- TB = W ( 2 - 3k ) / 6
Find the set of possible values of k for which both ropes remain taut.
( A non-uniform beam AD has weight W newtons and length 4 m. It is held in equilibrium in a horizontal position by two vertical ropes attached to the beam.
The ropes are attached to two points B and C on the beam, where AB = 1 m and CD = 1 m, as shown in Figure 7.
The tension in the rope attached to C is double the tension in the rope attached to B.
The beam is modelled as a rod and the ropes are modelled as light inextensible strings.
( TB = W ( 2 - 3k ) / 6 )
A small load of weight kW newtons is attached to the beam at D. The beam remains in equilibrium in a horizontal position.
The load is modelled as a particle. )
( Figure shows a rod AD ) ( AD is 4m apart ) ( Ropes are attached to B and C ) ( B is 1m from A ) ( C is 1m from B ) ( Weight of beam is WN ) ( D has a weight of kWN ) ( BW is 4 / 3m ) ( CW is 2 / 3m ) ( Ropes have a force of R and 2R for B and C ) ( BC is 2m apart )
- TB => 0
- W ( 2 - 3k ) / 6 <= 0
- 2 - 3k <= 0
- 3k <= - 2
- k <= 2 / 3
- 0 < k <= 2 / 3
A uniform rod AB has length 2 m and mass 50 kg.
The rod is in equilibrium in a horizontal position, resting on two smooth supports at C and D, where AC=0.2 metres and
DB=xmetres, as shown in Figure 8.
Given that the magnitude of the reaction on the rod at D is twice the magnitude of the reaction on the rod at C.
The support at D is now moved to the point E on the rod, where EB = 0.4 metres.
A particle of mass m kg is placed on the rod at B, and the rod remains in equilibrium in a horizontal position.
Given that the magnitude of the reaction on the rod at E is four times the magnitude of the reaction on the rod at C,
Find the value of m.
( Figure shows a rod AB ) ( AB is 2m apart ) ( Supports C and E are present ) ( C is 0.2m from A ) ( E is 0.4m from B ) ( Rod was a weight of 50kg ) ( Uniform rod, so 50kg is 1m from A and B ) ( Midpoint to E is 0.6m ) ( Midpoint to C is 0.8m ) ( Mkg is at B )
- R + 5R = 50g + mg
- 5R = 50g + mg
- R = 50g + mg / 5
A:
- ( 0.2 x R ) + ( 1.6 x 4R ) = ( 1 x 50g ) + ( 2 x mg )
- 0.2R + 6.4R = 50g + 2mg
- 6.6R = 50g + 2mg
- 6.6( 50g + mg / 5 ) = 50g + 2mg
- 330g + 6.6mg / 5 = 50g + 2mg
- 330g + 6.6mg = 250g + 10mg
- 3.4mg = 80g
- m = 400 / 17
A beam AB has length 15m.
The beam rests horizontally in equilibrium on two smooth
supports at the points P and Q, where AP = 2m and QB = 3m. When a child of mass 50 kg stands on the beam at A, the beam remains in equilibrium and is on the point of tilting
aboutP.
When the same child of mass 50kg stands on the beam at B, the beam remains in equilibrium and is on the point of tilting about Q. The child is modelled as a particle and the beam is modelled as a non-uniform rod.
Find the mass of the beam.
( Beam AB is 15m apart ) ( Beam has a weight of Mkg ) ( Supports P and Q are present ) ( AP is 2m apart ) ( QB is apart by 3m ) ( A has a weight of 50kg, which is labelled 1 ( by me ) ) ( B has a weight of 50kg, labelled 2 ( by me ) ) ( A to Mkg is xm )
P: ( Tilting at P )
- ( ( x - 2 ) x mg ) = ( 2 x 50g )
- xmg - 2mg = 100g
Q: ( Tilting at Q )
- ( 3 x 50g ) = ( 12 - x ) x mg )
- ( 12, is from 15 - 3, from B )
- 12mg - xmg = 150g
- ( Forms a simultaneous equation )
- xmg - 2mg = 100g
- xmg + 12mg = 150g
- ( Added )
- 10mg = 250g
- m = 25kg
