A. You drag a suitcase of mass 19 kg with a force of F at an angle 41.3 degrees with respect to the horizontal along a surface with kinetic coefficient of friction 0.54. The acceleration due to gravity is 9.8 m/s^2. What is the normal force on the suitcase? Answer in units of N.
So I understand that Normal force should be mgcos@, which gets me about 139.9N, but the program is telling me that this is wrong.
B. If you are accelerating the suitcase with acceleration 0.684 m/s^2, what is F? answer in units of N.
For this one, I think I need the normal force in order to find the frictional force, in order to solve for F, because Fcos@ should be greater than the frictional force and the x component of the normal force combined by about 13 N (mass of 19 times the given acceleration of 0.684.). If I'm wrong in there anywhere, please tell me.
A. m*g = 19kg * 9.8N/kg = 186.2 N. = Force of the suitcase.
u = a/g,
a = u*g = 0.54 * 9.8 = 5.29 m/s^2
Fn = mg-F*sin41.3 = 186.2-0.66F = Normal
force.
F*Cos41.3-.54(186.2-0.66F) = 19*5.29
0.751F - 100.5 + 0.356F = 100.5
1.107F = 201
F = 181.6 N. = Force applied.
Fn = 186.2 - 0.66*181.6 = 66.34 N. = Normal force.
B. F*Cos41.3-u*Fn = m*a
0.751F - 0.54*66.34 = 19*0.684 = 13
0.751F - 35.82 = 13
0.751F = 48.82
F = 65 N.
NOTES:
1. The surface is assumed to be horizontal.
2. The 43.1o is the angle at which the
suitcase is being pulled.
A. To calculate the normal force on the suitcase, we can use the equation:
Normal force (N) = mass (m) × acceleration due to gravity (g) × cosine(angle)
Plugging in the values:
Normal force = 19 kg × 9.8 m/s^2 × cos(41.3°)
Normal force = 19 kg × 9.8 m/s^2 × 0.754
Normal force ≈ 141.33 N
So, the normal force on the suitcase is approximately 141.33 N.
It seems that you calculated the normal force correctly. Please double-check your calculations.
B. To find the force (F) required to accelerate the suitcase, we need to consider the net force acting on it.
Net force = frictional force + force (F) × cosine(angle)
The frictional force can be calculated using:
Frictional force = coefficient of friction × normal force
Plugging in the values:
Frictional force = 0.54 × 141.33 N
Frictional force ≈ 76.34 N
Now, we know the net force required to accelerate the suitcase:
Net force = mass × acceleration
Net force = 19 kg × 0.684 m/s^2
Net force ≈ 13.00 N
Since net force = force (F) × cosine(angle) + frictional force, we can rearrange the equation to solve for force (F):
Force (F) = (net force - frictional force) / cosine(angle)
Force (F) = (13.00 N - 76.34 N) / cos(41.3°)
Force (F) ≈ -94.07 N
Note: The negative sign indicates that the force is in the opposite direction of motion.
So, the force (F) required to accelerate the suitcase is approximately -94.07 N.
A. To find the normal force on the suitcase, you are correct that it can be calculated using the equation N = mgcosθ, where m is the mass of the suitcase, g is the acceleration due to gravity, and θ is the angle with respect to the horizontal. Given that the mass of the suitcase is 19 kg, and the angle is 41.3 degrees, we can calculate the normal force as follows:
N = (19 kg)(9.8 m/s^2)cos(41.3 degrees)
N ≈ 137.29 N
So, your calculation of 139.9 N is very close, and it seems like a rounding error might have caused the program to mark your answer as incorrect. If you round the value to the nearest whole number, it should be accepted as the correct answer.
B. Now, to find the force (F) required to accelerate the suitcase with a given acceleration of 0.684 m/s^2, you are correct that you need to consider the frictional force and the x-component of the normal force.
First, let's calculate the frictional force (f) using the equation f = μN, where μ is the kinetic coefficient of friction (0.54 in this case) and N is the normal force we calculated earlier:
f = (0.54)(137.29 N)
f ≈ 74.23 N
Next, we need to find the total force in the x-direction. This force is a combination of the horizontal component of the force applied (Fcosθ) and the frictional force (f):
Fcosθ + f > mass × acceleration
Fcosθ > mass × acceleration - f
Fcosθ > (19 kg)(0.684 m/s^2) - 74.23 N
Now, solving for F:
F > [(19 kg)(0.684 m/s^2) - 74.23 N] / cos(41.3 degrees)
By plugging in the values, we can calculate F:
F > [(19 kg)(0.684 m/s^2) - 74.23 N] / cos(41.3 degrees)
F > 7.91 N / 0.753
F > 10.50 N
Therefore, the force required to accelerate the suitcase with an acceleration of 0.684 m/s^2 is approximately 10.50 N.