Question

This assignment is set up for sequential assessment Complete each question and submit the answer before moving on to the next question Correct answers for each question will be made available once the maximum number of attempts have been made for each submission length of Ladler -7.om Static Equilibrium - Ladder Comprehensive As shown in the figure, the top of a 7.0 m long ladder (mass 2 -20 kg) leans against a frictionless wall. The bottom of the ladder rests on a rough floor, with 3.2 m from the wall. The distance from the ladder's center of mass to its bottom end is 2.5 m (note: this ladder is not uniform, therefore its center of mass is not at the midpoint). A man of mass 1 = 87 kg stands on the middle of the ladder. The diagram also shows all the relevant forces (a) Find h(the height of the top of the ladder) (in meters) in the drawing. Use your knowledge of geometry. 6.23 Correct (100,0%)

please answer unsolved questions

Answer 1

We: 20kg Wi=87kg AK L=7m A 3.5mit s's com -floor 3.2 m (a) from Pythagorean theorem AB = AC? - BC2 h = 72 – 3.22 h=V72 – 3.22 h= 6.23m AABC and ADEC are similar therefore

EE E RE = 32 di = 3.2 *3.5 = 1.6m AABC and AFGC are similar therefore GE = RE en = 3:2 d2 = 3.2 $2.5 = 1.143m N=W +W2 N = (87) X (9.8) + (20) x (9.8) N = 1048.6N

We: 20kg han i W ,=87kg L:7m 3. 5m man D oder's com floor EG-4 ad -3. 2m - pivote (c) Batrom end of the ladder Since normal is passing through rotational axis therefore its torque is zero T= XÑ r=0 T=ON.m friction is also passing through rotational axis therefore its torque is zero T = 7 xf r=0 I=ON.m magnitude of torque produced by weight of man is

Ti = dW T1 = (1.6) X (87) X (9.8) T = 1364.16N.m magnitude of torque produced by weight of ladder T2 = d. T2 = (1.143) x (20) (9.8) t2 = 224.028 Nm lever arm of force P is h = 6.23m Torque by P tp = hP Tp = 6.23P In rotational equilibrium Tnet = 0 11 + 2 = 1p 1364.16 +224.028 = 6.23P P= 1364.16 + 224.028 6.23 P= 254.926N (f) Fx, net = 0 f-P=0 f =P f = 254.926N