Moment exam questions

Folie 1

• 
• Bildunterschrift: The mass of the ruler is 160g

Answer 1....

• 50-23 = 27cm , 88-23 = 65cm -> 0.27m and 0.65m. 160g=0.16kg , 190g =0.19kg , 950g=0.95kg
• (9.81 x 0.16) x 0.27 + (9.81 x 0.19) x 0.65 = 1.635 (this is the CW moment, and as we are in equilibrium this is also the AC moment)
• 1.635 / (9.81 x 0.95) = 0.1755m (this is where the mass is located, as we are diving the AC moment by force to get perpendicular distance)
• 0.23 - 0.1755 = 0.05m (this is the distance away from the pivot and therefore our answer)

Folie 3

• 
• Bildunterschrift: (The moment is in equilibrium) , 740g mass is at 92cm

Answer 2....

• 92-81 = 11cm , 81-50=31cm -> 11cm=0.11m , 31cm=0.31m 85g = 0.085kg , 210g = 0.21kg , 740g = 0.74kg
• CW moment: (9.81 x 0.74) x 0.11 , and as we have an equilibrium we can minus one AC moment to get the other AC moment)
• So: (9.81 x 0.74) x 0.11 - (9.81 x 0.085) x 0.31 = 0.54Nm
• Now we divide this by its weight to get perpendiular distance. So 0.54 / (9.81 x 0.21) = 0.2621
• Finally we minus this from the pivot distance to get distance way from pivot. So 0.81-0.2621 = 0.55m

Folie 5

• 
• Bildunterschrift: Also find the force of K

Answer 3

• Firstly find the distance from K to the centre of the beam by doing 1.5m-0.4m to get 1.1m.
• Now do: (22 x 9.81) x 1.1 + (52 x 9.81) x 1.7 to get 1104.606Nm. (this step is simply finding the total CW moment from K)
• Finally we divide our answer by the distance of KL (1.7m + 0.5m = 2.2m). So 1104.606 / 2.2 = 502N (force acting at L)
• To find the force of K we simply add our downward forces (22 x 9.81 + 52 x 9.81) to get 725.94 and take away our answer of 502 to get 224N to 3sf.

Folie 7

Answer 4

• 1. Weight (down from the 25N force) and reaction (at the hinge)
• 2. We need to imagine this as a big right angled triangle, which has another right angled triangle inside of it which only goes up to the 25N force. As we don't have any perpendicular distances, we do: cos60 x 0.9 to get 0.45m and cos60 x 0.35 to get 0.175m.
• Now we have these we know that: V x 0.45m = 0.175m x 25N (as equilibrium and CW=AC) so V=0.175 x 25 / 0.45 , V= 9.7N
• 3. As we have an equilibiuim, we know that the upward forces must equal the downward force, so we can say: 9.7N + R -25 = 0, so therefore 25-9.7 = R , so R = 15.3N

Folie 9

Answer 5

• Firstly for part a, we find the AC moment of the 450N 40cm away from the pivot by doing 40cos30 x 450 (to get 15588.5Ncm)
• Next we find the way the vertrical distance of F, so we do 25sin15 (to get 6.47cm)
• Finally we 450 x 40cos30 / 25sin15 to get 2409N. This is part i. For part ii, replace 40cos30 with 40cos70 to get 951N
• For part b we need to recognise and state:
• 1. Larger force with smaller angle
• 2. AC moment large as perpendicular distance to pivot is large
• 3. The best position would be to bend knees and keep the load close to your body.
• (essentially we are keeping the distance as short as possible).