[" Far the cqulibrium condtion shown,the cards are strang enaugh to wittstand a maximum tension "100],[" N,What is the lagest value of "W" ( hin Newoon) that can te suspended "],[qquad (mu)/(53^(@))],[qquad [" A "],[53^(@)]]

For the equilibrium situation shown, the cords are strong enough to with stand a maximum tension 100 N. what is the largest value of W (in N) that they can support as shown ?

A ball of mass 0.5 kg is attached to a light string of length 2m. The ball is whirled on la horizontal smooth surface is a circle of rdius 2 m as shown in fig. If the string can withstand tension up to 100 N, what is the maximum speed at which the bal can be whirled.

Passage IV) Angular frequency in SHM is given by omega=sqrt(k/m) . Maximum acceleration in SHM is omega^(2) A and maximum value of friction between two bodies in contact is muN , where N is the normal reaction between the bodies. In the figure shown, what can be the maximum amplitude of the system so that there is no slipping between any of hte blocks?

Passage IV) Angular frequency in SHM is given by omega=sqrt(k/m) . Maximum acceleration in SHM is omega^(2) A and maximum value of friction between two bodies in contact is muN , where N is the normal reaction between the bodies. In the figure shown, what can be the maximum amplitude of the system so that there is no slipping between any of hte blocks?

Passage IV) Angular frequency in SHM is given by omega=sqrt(k/m) . Maximum acceleration in SHM is omega^(2) A and maximum value of friction between two bodies in contact is muN , where N is the normal reaction between the bodies. In the figure shown, what can be the maximum amplitude of the system so that there is no slipping between any of hte blocks?

Two block tied with a massless string of length 3m are placed on a rotating table as shown. The axis of rotation is 1m from 1kg mass and 2m from 2kg mass. The angular speed omega=4rad//s . Ground below 2kg block is smooth and below 1kg block is rough. (g=10m//s^(2)) (a) Find tension in the string, force of friction on 1kg block and its direction. If coefficient of friction between 1kg block and groung is mu=0.8 Find maximum angular speed so that neither of the blocksd slips. (c) If maximum tension in the string can be 100N , then maximum angular speed so that neither of the blocks slips.

Two block tied with a massless string of length 3m are placed on a rotating table as shown. The axis of rotation is 1m from 1kg mass and 2m from 2kg mass. The angular speed omega=4rad//s . Ground below 2kg block is smooth and below 1kg block is rough. (g=10m//s^(2)) (a) Find tension in the string, force of friction on 1kg block and its direction. (b) If coefficient of friction between 1kg block and groung is mu=0.8 . Find maximum angular speed so that neither of the blocksd slips. (c) If maximum tension in the string can be 100N , then maximum angular speed so that neither of the blocks slips.

After receiving the lamp, Aladdin and Abu are on their way to get out from the cave when they found a statue of Sir Hussain, one of the most powerful Royal General to walk on Earth. Below the statue is written a real life problem that Sir Hussain solved and in order to get out of the cave one must solve this riddle. Aladdin out of options starts reading the riddle- Once upon a time in the Kingdom far far away lived Sir Hussain, the chief Royal General. He was very proud of his men and he liked to invite the King to come and watch drill exer- cises which demonstrated the fighting techniques and tactics of the squad he was in charge of. But time went by and one-day Sir Hussain had a major argument with an old witch (there were rumours that the argument occurred after the general spoke badly of the witch flying techniques. That seemed to hurt the old witch very deeply). As the result of the argument, the witch put a rather strange curse upon the general. It sounded all complicated and quite harmless: "If the squared distance between some two soldiers equals to 5, then those soldiers will conflict with each other!" The drill exercises are held on a rectangular n × m field, split into nm square 1 × 1 segments for each soldier. Thus, the square of the distance between the soldiers that stand on squares (x_(1), y_(1)) and (x_(1),y_(2)) equals exactly (x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2) . Now not all nm squad soldiers can partici- pate in the drill exercises as it was before the old witch curse. Unless, of course, the general wants the soldiers to fight with each other or even worse. For example, if he puts a soldier in the square (2, 2), then he cannot put soldiers in the squares (1, 4), (3, 4), (4, 1) and (4, 3) — each of them will conflict with the soldier in the square (2, 2). Find the sum of maximum number of soldiers that can be simultaneously positioned on this field for each of the following cases i) 53 xx 81 II) 2 xx 103 III) 1 xx 104

A system of two blocks is shown in figure. Friction coefficient between 5 kg and 10 kg block is mu=0.6 and between 10 kg and ground is mu=0.4 What will be the maximum value of force F applied at the lower block so that 5 kg block does not slip w.r.t. 10 kg . (g=10m//sec^(2)). The force applied at the upper block is having fixed magnitude of 80 N (both forces start to act simultaneously)