Force = mass * acceleration
10 N - 2 N = 20 kg * acceleration
8 N = 20 kg * acceleration
8 / 20 = acceleration
2/5 m/s^2 = acceleration
Answer:
The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.
C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'
Explanation:
See attached picture.
Complete Question
Part of the question is shown on the first uploaded image
The rest of the question
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.
Answer:
The net force exerted on the third charge is
Explanation:
From the question we are told that
The third charge is 
The position of the third charge is 
The first charge is 
The position of the first charge is 
The second charge is 
The position of the second charge is
The distance between the first and the third charge is


The force exerted on the third charge by the first is

Where k is the coulomb's constant with a value 
substituting values
The distance between the second and the third charge is


The force exerted on the third charge by the first is mathematically evaluated as
substituting values

The net force is
substituting values

Answer:
The answer to your question is Decrease
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Hope this helps! :)