When a man in a cage whose weight is 1000N pulls a rope that is attached on a cage around a fixed pulley,the force he exerts on the floor of the cage is 450N ,if the cage weigh 250N ,find the acceleration
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I don't get the picture...a man in a cage pulls a rope attached to the cage..
Picture it like this:the rope is attached on top of the cage then from there it passes around a fixed pulley so the man in the cage is pulling the rope down from the pulley.PLEASE HELP ME Bobpursley
The weight of cage + man is 1250 N and their combined mass is
Mtotal = Wtotal/g = 127.6 kg.
The man and cage accelerate at the same rate, a. Let F be the rope tension force. Consider all forces on the man, including weight and the upward floor force. The man's mass is 102 kg
F - Wman + 450 = F - 550 = 102*a
Now consider all exernal forces on man and cage together:
F -1250 = 127.6 a
Combining the two results in
700 = -25.6 a
a = -0.0366 m/s^2
Mtotal = Wtotal/g = 127.6 kg.
The man and cage accelerate at the same rate, a. Let F be the rope tension force. Consider all forces on the man, including weight and the upward floor force. The man's mass is 102 kg
F - Wman + 450 = F - 550 = 102*a
Now consider all exernal forces on man and cage together:
F -1250 = 127.6 a
Combining the two results in
700 = -25.6 a
a = -0.0366 m/s^2
I think u have to also take
T-f-Wchair=McA
T-f-Wchair=McA
U did a mistake
Wtotal/g=127.55
Wtotal/g=127.55
1 answer