Answer:
shown in the attachment
Explanation:
The detailed step by step and necessary mathematical application is as shown in the attachment.
Fnet=(115+106)-186= 34 N
mass=Force/g= 186N/9.8m/s^2 = 18.98 kg
a=fnet/mass => 34N/18.98kg = 1.79 m/s^2
so A= 1.8m/s^2
Answer
The answer and procedures of the exercise are attached in the following archives.
Explanation
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
The percentage of the weight supported by the front wheel is A= 19.82 %
Explanation:
From the question we are told that
The center of gravity of the plane to its nose is 
The distance of the front wheel of the plane to its nose is 
The distance of the main wheel of the plane to its nose is 
At equilibrium the Torque about the nose of the airplane is mathematically represented as

Where m is the mass of the airplane
is the weight of the airplane supported by the main wheel
So

Substituting values
Now the weight supported at the frontal wheel is mathematically evaluated as

Substituting values
Now the weight of the airplane is = mg
Thus percentage of this weight supported by the front wheel is
19.82 %