Why is the downward force on 1kg in a force pulley system 60N?

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In a force pulley system, the downward force on a 1 kg mass is calculated as 60N due to the combined effects of gravity and tension in the string. The weight of the 1 kg mass is 10N, but when considering the entire system, the acceleration of 5 m/s² results in additional tension forces. The tension in the string connecting the 5 kg mass and the 1 kg mass must also be accounted for, leading to a total downward force of 60N on the 1 kg mass. Understanding these forces requires analyzing the entire system rather than just the individual mass. This illustrates the complexity of forces in a pulley system.
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If a free body diagram is constructed for 1kg, one might be inclined to draw an arrow downwards with only the weight of 10N. However, the downward, non-net, force on 1 kg is (1+5) x 9.8. Why is this the case? It seems very counterintuitive.
 
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aspodkfpo said:
Homework Statement:: n/a
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View attachment 267189

If a free body diagram is constructed for 1kg, one might be inclined to draw an arrow downwards with only the weight of 10N. However, the downward, non-net, force on 1 kg is (1+5) x 9.8. Why is this the case? It seems very counterintuitive.
It is not true, you misunderstood something. There are to forces acting on the 1 kg mass, its weight, 1*g downward, and the tension T1 in the string, upward.
 
ehild said:
It is not true, you misunderstood something. There are to forces acting on the 1 kg mass, its weight, 1*g downward, and the tension T1 in the string, upward.
Consider the entire system.
(1+5) x 10 is the force acting on the entire system. This means that the system travels with an acceleration of 5ms^-2. The tension in the string to the right of 6kg is 30N. 5kg has to have a net force of 25N, so between 5kg and 1kg, string is at 55N. 1kg has to travel at 5ms^-2, so downwards on 1kg is 60N.

I don't quite see how it is intuitive to realize it is 60N pointing in the downward direction on the 1kg.
 
Cut the string between the two hanging masses and remove the smallest mass.
How much force your hand should pull down the section of remaining string hanging from mass 5kg to keep the original tension in longer string?
 
aspodkfpo said:
Homework Statement:: n/a
Relevant Equations:: n/a

View attachment 267189

If a free body diagram is constructed for 1kg, one might be inclined to draw an arrow downwards with only the weight of 10N. However, the downward, non-net, force on 1 kg is (1+5) x 9.8. Why is this the case? It seems very counterintuitive.
I can think of two forces acting on the 1-kg mass: The tension in the string (up) which is exerted by the string and gravity (down) which is exerted by the Earth. Their sum is the net force and that is directed down. What force is the "downward non-net force" and what entity exerts it?
 
aspodkfpo said:
Consider the entire system.
(1+5) x 10 is the force acting on the entire system. This means that the system travels with an acceleration of 5ms^-2. The tension in the string to the right of 6kg is 30N. 5kg has to have a net force of 25N,
correct so far, but wrong after.
aspodkfpo said:
so between 5kg and 1kg, string is at 55N. 1kg has to travel at 5ms^-2, so downwards on 1kg is 60N.

I don't quite see how it is intuitive to realize it is 60N pointing in the downward direction on the 1kg.
There are three forces acting on the 5 kg mass. The tension in the upper string, 30 N upwards
the weight, 50 N, downwards
and the tension T1, connecting 5kg an 1 kg. It acts downward. The 5 kg mass accelerates downward with 5m/s2, so the net force acting on it is T1+50-30=25 N. what is the value of T1 then?
 
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