The system in the image below is at equilibrium. The smooth rod has a mass of 8.00 kg, and a centre of mass at point G, which is halfway along the length of the rod. You can neglect the mass of the rope, which attaches to a hook on the wall. Determine the distance along the wall between the hook and the rod. (g = 9.81m/s^2)

Here is some trigonometry information that may help you (note: you do not have to use this):

sin(180 - theta) = sin(theta)

Cosine law: a^2 = b^2 + c^2 -2bc*cos(alpha)

Transcribed Image Text:

(10 marks) The system in the image below is at equilibrium. The smooth rod has a mass of 8.00 kg, and a centre of mass at point G, which is halfway along the length of the rod. You can neglect the mass of the rope, which attaches to a hack on the wall. Determine the distance along the wall between the hook and the rod. (g-9.61m/s 21 Here is some trigonometry information that may help you (nate: you do not have to use thisk sin/180-theta)=sin(theta) Cosine law: a 2 = bx2-02-2bconsjalaha) (For full marks you need to show some work) A 5.00 m C 4.00 m G XXXXXXXXXxxxxxxxxxx B ? (10 marks) The system in the image below is at equilibrium. The smooth rod has a mass of 8.00 kg, and a centre of mass at point G, which is halfway along the length of the rod. You can neglect the mass of the rope, which attaches to a hack on the wall. Determine the distance along the wall between the hook and the rod. (g-9.61m/s 21 Here is some trigonometry information that may help you (nate: you do not have to use thisk sin/180-theta)=sin(theta) Cosine law: a 2 = bx2-02-2bconsjalaha) (For full marks you need to show some work) A 5.00 m C 4.00 m G XXXXXXXXXxxxxxxxxxx B ?

Answer rating: 100% (QA)

Calculation T A W moment MB0 B F... View the full answer