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. 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.
1 answer
1 answer