Dynamics Question: Finding Fa using mass and acceleration

In summary, the force available to accelerate the object is equal to the sum of all the forces acting on the object. This equation can be rearranged to find the applied force, which is equal to the mass of the object multiplied by the sum of the acceleration and the gravitational force. The correct solution for finding the applied force in this scenario is 15 (8.8 + 9.81) = 279 N.
  • #1
Apollinaria
82
0
Hello again forum :smile: I am still studying for that unit test and am still having problems with dynamics... I've attached the problem with the calculations/solutions. Basically, I don't understand why I'm getting the wrong answer. Probably b/c idk what the heck I'm doing :eek:

Homework Statement



15kg object thrown into the air. Initial acceleration is 8.8m/s^2. Find applied force.


Homework Equations



Please see attachment.

The Attempt at a Solution



Please see attachment. Btw, I realize Fa needs to be positive.

Please advise regarding correct way of solving?.. Thank you for your time once again:smile:

EDIT: I found my mistake. However, I still don't understand why "that" is the correct solution.
My Fg value had to be negative;
Fa= (15kg x 8.8m/s^2) - (-147.15)
Fa= 279.15
 

Attachments

  • Dynamics.jpg
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  • #2
The force available to accelerate the object (Fnet) is equal to the sum of all the forces acting on the object.

Fnet = FA + (-Fg)
or
Fnet = FA - Fg

Fg is negative because it's in the oposite direction to FA.

Rearrange..

FA = Fnet + Fg

= ma + mg
= m (a+g)
= 15 (8.8 + 9.81)
= 279 N
 
  • #3
You have done everything right, you simply made a mistake while solving the equations.
You wrote:
FA-Fg=ma
FA=ma-Fg. <--- This is wrong. It should be FA=ma+Fg. :wink:
 
  • #4
Well spotted. I looked at his working and didn't see that error.
 
  • #5
CWatters said:
The force available to accelerate the object (Fnet) is equal to the sum of all the forces acting on the object.

Fnet = FA + (-Fg)
or
Fnet = FA - Fg

Fg is negative because it's in the oposite direction to FA.

Rearrange..

FA = Fnet + Fg

= ma + mg
= m (a+g)
= 15 (8.8 + 9.81)
= 279 N

Thank you Watters :biggrin: That organizes it a little better lol. In regard to your last post, I am very much female by the way :wink:
 
  • #6
Pranav-Arora said:
You have done everything right, you simply made a mistake while solving the equations.
You wrote:
FA-Fg=ma
FA=ma-Fg. <--- This is wrong. It should be FA=ma+Fg. :wink:

Eek! So sad that I didn't notice that... I'm sorry you had to point it out but thank you once again :biggrin:
 

FAQ: Dynamics Question: Finding Fa using mass and acceleration

What is the formula for finding force (Fa) using mass and acceleration?

The formula for finding force (Fa) using mass (m) and acceleration (a) is Fa = m x a. This formula is known as Newton's second law of motion.

How do you calculate the force (Fa) when given the mass and acceleration?

To calculate the force (Fa) when given the mass (m) and acceleration (a), simply multiply the mass by the acceleration. This will give you the value of the force in Newtons (N).

What units are used to measure force (Fa), mass, and acceleration?

The unit of force is Newtons (N), the unit of mass is kilograms (kg), and the unit of acceleration is meters per second squared (m/s2). It is important to use the correct units when calculating force using mass and acceleration.

Can force (Fa) be negative when using mass and acceleration?

Yes, force (Fa) can be negative when using mass and acceleration. This indicates that the force is acting in the opposite direction of the acceleration. For example, if an object is moving to the left with a negative acceleration, the force acting on it will be in the positive direction.

How does the mass and acceleration affect the force (Fa) in an object?

The mass and acceleration of an object directly affect the force (Fa) acting on it. The greater the mass of an object, the more force is needed to accelerate it. Similarly, the greater the acceleration of an object, the more force is needed to maintain that acceleration.

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