# Physics Help?

A woman at an airport tows her 17.6 kg suitcase behind her as she runs to make her connecting flight. She accelerates at 1.20 m/s2 as she pulls with 49.5 N on the suitcase strap at an angle θ above the horizontal, as shown in the diagram.(Figure 1) The frictional force on the suitcase is 8.70 N .

1) What is the value of θ, the angle the suitcase strap makes with the horizontal?

2)What is the magnitude of the normal force, n, that the airport floor exerts on the suitcase?

3)Calculate the coefficient of kinetic friction between the suitcase and the airport floor.

Relevance
• 1) since a = F cos(theta)/m - f/m

ie., 1.2 = 49.5 cos(theta)/ 17.6 - 8.7 /17.6

ie., cos(theta) = 1.7x17.6/49.5 = 0.6044 = cos(53)

hence theta = 53 degrees.

2) normal force N = mg - F sin(53) = 176 - 49.5x0.7986 = 176-31.6 = 144.4 N.

3) coefficient of friction= f / N = 8.7 / 144.4 = 0.06

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• (a) The pull of 49.5 N at θ above the horizontal will have two components effecting the motion of the suitcase i.e. vertical component (Fy = 49.5 sinθ) acting against the weight of the suitcase & horizontal component (Fx = 49.5 cosθ) acting horizontal to provide the motion against the friction force(Ff).

Thus N = W - Fy = 17.6 x 9.8 - 49.5sinθ -----------------(i)

& F(net) = Fx - Ff

=>ma = 49.5 cosθ - 8.70

=>17.6 x 1.2 = 49.5 cosθ - 8.70

=>cosθ = 0.60 = cos 52.96

=>θ = 52*96'

(b) By (i):

=>N = 132.97 N

(c) By Ff = µk x N

=>µk = 8.70/132.97= 0.065

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• general equation :

(m*g+49.5*sin Θ )*μ +m*a = 49.5*cos Θ

since (m*g+49.5*sin Θ )*μ = friction force FF = 8.70 N, then :

8.70+17.6*1.20 = 49.5*cos Θ

cos Θ = (8.70+17.6*1.20) / 49.5 = 0.60 ; sin Θ = 0.80

angle Θ = arccos 0.6 = 53.1°

reaction force RF = m*g-49,5*sin 53.1° = (17.6*9.806-49.5*0.8) = 133 N up

friction coefficient μ = FF/RF = 8.70/133 = 0.065

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• Need the diagram to correctly work the problem

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