Recent content by khuko

Thanks, i now see that you must do mg/5 for the answer. Thank you very much!. How do i must learn to solve some exercises like this? caus i don't understand the essential bout this exercise

There are two forces that are pulling the pulley up, both are T1 and one goes down. That is the weight of the object (man+chair). therefore the equation is 2*T1-mg =>T1=mg/2 And if I substitute this in 2*T2=T1 We get for T2= mg/4 ow no. just saw that the bottom pulley haves two tensions...

Hmm The Two forces that are pulling the pulley upwards are T2. The first T2 goes to the ceiling and haves a normalforce called T2' who's going back to the rope 2 (and pulls the pulley upwards). The force that pulls the pulley downwards is the normalforce of T1 called T1' . We have also...

Allright. the middle pulley is the "center" point of all the tensions from the ropes. everythings comes to this point (correct ?). T1 is the tension in rope 1, T2 is the tension in rope 2 There's no force in the x-direction. y-direction : The net force of the pulley must be zero cause...

i must be -"T". So the equation for the rope 2 => T=mg/2 and for rope 1 => -T=-mg/2. (The opposite ). But a negative force doesn't exist :s But i don't understand why the answer is 137,2 N. Cause my answer would be for rope 2 = 343,35

If he pulls on rope 2 , then will the pulley follow and the chair go up. So it's related to each other or else the two ropes where indepented and the weight (and tension) where the same for each rope. ?

Homework Statement Calculate the force that the man (weight 70 kg) should exert on the rope to stay in balance. (answer = 137,2 N) [PLAIN]http://img651.imageshack.us/img651/7682/oefkoen.png Homework Equations Like I see, there are two ropes who are in equilibrium. In each rope...