PDA

View Full Version : Static Equilibrium Questions.


physicshelpkanha
Oct 22, 2007, 06:08 PM
Question 1
Where must an 800N weight be hung on a uniform, 100N board so that a girl at one end supports 1/3 as much as the woman at the other end?:confused:

Question 2
An 8.00m ladder leaning up against a wall makes an angle of 60 degrees with the ground. A 60.0 kg worker climbs 3/4 of the way up ladder. A rope at the base of the ladder is attached to the wall for extra support. Between the ladder and the ground Satic mu is 0.05. The wall is very icy and frictionless. The mass of the ladder is 10.0 kg. What tension force (if any) will be required in the rope in order to keep the ladder from slipping?:confused:

terryg752
Oct 22, 2007, 11:20 PM
1 .Suppose board is AB

Suppose 800 N is hung distance y from B.

Suppose mid point of board is distance x + y from B

Length of board = L = 2(x + y)

100 N acts at mid point of board.

Total weight to support is = 800 + 100 = 900

support at A = 225, Support at B = 675

(because support at B = 3 times support at A, and total = 800 + 100 = 900)

100 acts at the mid point.

Take moments about B

225 L = 100 L/2+ 800 y

175 L = 800 y


y = 175/800 L

= 7/32 L

Hence 800 N must be hung distance 7/32 L from B (where L is length of board)

2. Total weight downward = 70g (g = gravity)

Hence total upward force = 70 g

Hence Normal Reaction = N at the ground end A of ladder = 70g

Hence force of friction .05 times 70 g

= 3.5g

Let Tension = T

Take moments about A (where ladder touches wall)

Let the distance of B from wall = x

Let the wall end of ladder be distance y from ground

(T + Friction) y + 10g x/2 + 60 g 1/4 x = N x

(T + 3.5 g) y + 5xg + 15 xg = 70 xg

y/x = tan 60 Hence y = x tan 60

(T + 3.5g) x tan 60 + 20 xg = 70gx

(T + 3.5g) x tan 60 = 50 gx

T + 3.5g = 50 g cot 60

T = 50 g cot 60 - 3.5g

Substitute values of g and Cot 60 to get tension.

physicshelpkanha
Oct 23, 2007, 04:51 PM
Thanks allot for ur help
i really appreaciate it...
Thank You!