Three forces on a box in different directions what is the acceleration

In summary: The same mistake is repeated for Fbx. Also, for Fb and Fc, the angles should be 120 and 240 degrees respectively. In summary, using the given forces and angles, we can calculate the components of each force and use the equations F = ma and Ff = μFn to find the net force and acceleration of the object with mass 40kg.
  • #1
Lucaburra
3
0

Homework Statement



Force A 250 degrees @ 200 N
Force B 150 degrees @ 300 N
Force C 50 degrees @ 400 N
mass = 40kg
mew = .14
what is acceleration

Homework Equations



Fa^2 = Fy net^2 x Fx net^2
Fn = Fg
Ff = Fn x mew
Fnet = Fa - Ff
F= ma


The Attempt at a Solution



Fax=Cos 20 (200) = -187.93
Fay = Sin 20 (200) = -68.40

Fbx = Sin 30 (300) = 150
Fby = Cos 30 (300) = -259.81

Fcx = Cos 50 (400) = 257.11
Fcy = Sin 50 (400) = 306.42

Fx net = 219.18 N
Fynet = -21.79 N

Fa^2 = Fxnet^2 + Fynet^2
Fa^2 = 219.18^2 + -21.79^2
Fa^2 = 48039.87 + -474.80
Fa^2 = 47565.07
Fa = 218.09 N

Fn = 9.81 m/s^2 x 40Kg = 392.40N

Ff = Fn (mew) = 392.40N (.14) = 54.94 N

Fnet = Fa - Ff = 218.09 - 54.94 = 163.15 N
a = F/m = 163.15/40 = 4.08 m/s

I'm just not sure I've done this correct
 
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  • #2
The components of Fa and Fb are not computed correctly. BTW, you could make your life easier by choosing x to coincide with one of the forces, say Fc, then Fcx = 400 N and Fcy = 0, and the angles of the other forces are relative to Fc, i.e., 200 and 100.
 
  • #3
Sorry voko I don't follow. I drew my x,y graph and entered the force vectors. I then used the angles relative to the x axis, (as the text says) hence F ax = Cos 20 and F bx = Sin 30. I'm not sure about your meaning to "choose x to coincide with one of the forces." I'm in grade 11 and this is what we were taught to do.
 
  • #4
No problem, you can ignore this part. But the components of Fa and Fb are still not computed properly. For example, Fax = Fa cos 250 = -68.4 N.
 
  • #5
.



Your calculations and approach seem to be correct. However, it would be helpful to provide a diagram or a clear explanation of the direction of each force and how they are affecting the box. This will help in understanding the final acceleration value and its direction. Additionally, it would be beneficial to include units for the final acceleration value (m/s^2). Overall, your solution appears to be accurate and well thought out.
 

1. What are the three forces on a box in different directions?

The three forces on a box in different directions are usually the weight, which acts downwards towards the center of the Earth, and two horizontal forces, which act in opposite directions along the surface of the box.

2. How do these forces affect the acceleration of the box?

The acceleration of the box will depend on the magnitude and direction of the forces acting on it. If the horizontal forces are equal in magnitude and opposite in direction, they will cancel out and the box will experience a net force equal to its weight, resulting in a downward acceleration. If one horizontal force is stronger than the other, the box will experience a net force in the direction of the stronger force, resulting in an acceleration in that direction.

3. Is the acceleration of the box constant?

If the forces acting on the box remain constant, then the acceleration will also be constant. However, if the forces change in magnitude or direction, the acceleration will also change accordingly.

4. How can we calculate the acceleration of the box?

To calculate the acceleration of the box, we can use Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration (F = ma). By knowing the mass of the box and the net force acting on it, we can calculate the acceleration using this formula.

5. What other factors may affect the acceleration of the box?

Aside from the forces acting on the box, other factors that may affect its acceleration include the surface it is moving on (e.g. friction), the shape and weight distribution of the box, and any external forces (e.g. air resistance) that may be acting on it.

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