Asked by Alyyssa
At the airport, you pull a 15kg suitcase across the floor with a strap that is at an angle of 45 degrees above the horizontal. Find the normal force and the tension in the strap if the suitcase is pulled at a constant speed and the coefficient of kinetic friction is 0.36.
Answers
Answered by
Henry
Ws = M*g = 15 * 9.8 = 147 N. = Wt. of suitcase.
Fn = Ws-T*sin45 = 147 - T*sin45 = 147-0.707T = Normal force.
Fk=u*Fn=0.36*(147-0.707T) = 52.9-0.255T.
T*Cos45-Fk = M*a.
0.707T-(52.9-0.255T) = M*0.
0.707T-52.9+0.255T = 0.
0.961T = 52.9
T = 55 N.
Fn = 147 - 0.707*55 = 108 N.
Fn = Ws-T*sin45 = 147 - T*sin45 = 147-0.707T = Normal force.
Fk=u*Fn=0.36*(147-0.707T) = 52.9-0.255T.
T*Cos45-Fk = M*a.
0.707T-(52.9-0.255T) = M*0.
0.707T-52.9+0.255T = 0.
0.961T = 52.9
T = 55 N.
Fn = 147 - 0.707*55 = 108 N.
Answered by
Anonymous
You pull your 15 kg suitcase at the airport. What is the coefficient of kinetic friction (uk) between the case and the floor if you pull with a force of 50 N, and the suitcase accelerates at 0.52 m/s²
Answered by
A
.52
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