# M1 Exam Paper QuestionsWatch

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#1
From jan 02 M1, Q5

A heavy uniform steel girder AB has length 10m. A load of weight 150N is attached to the girder at A and a load of weight 250N is attached to the girder at B. The loaded girder hangs in equil in a horizontal position held by two vertical steel cables attached to the girder at the points C and D, where AC = 1m and DB = 3m. The girder is modelled as a uniform rod, the loads s particles and the cables as light inextensible strings. The tension in the cable at D is three times the tension in the cable at C.

(a) draw a diagram showing all the forces acting on the girder. could someone do that in paint so i can see if i got it right?

(b)find the tension in the cable at C
(c) the weight of the girder
(d) explain how you have used teh fact that the girder is uniform

and then Q8, from the same paper:

Two particles P and Q have masses 3m and 5m respectively. They are connected by a light inextensible string which passes over a small smooth light pulley fixed at the edge of a rough horizontal table. particle P lies on the table and particle Q hangs freely below the pulley. The coeff of friction between P and the table is 0.6 The system is realsed from rest with the string taut. For the period before Q hits the floor or reaches th pulley.
...

(b) find, in terms of g, the accel of Q
(c) find, in terms of m and g, the tension in the string

When Q has moved a distance h, it hits the floor and the string becomes slack. Given tha t P remains on the table during the subsequent motion and does not reach the pulley
(d), find, in terms of h, the distance moved by P after the string becomes slack until P comes to rest.

Thanks
0
13 years ago
#2
Diagram...
0
13 years ago
#3
T + 3T = 150 + W + 120
4T = 400 + W
W = 4T -400

At about point C: (4*W) - (6*3T) + (9 * 250) - (1 * 150) = 0

Sustitute this : W = 4T -400 into above equation. You will then find T.
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