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Question

A woman at an airport is towing her $$20.0kg$$ suitcase at constant speed by pulling on a strap at an angle $$\theta$$ above the horizontal (Fig. above). She pulls on the strap with a $$35.0N$$ force, and the friction force on the suitcase is $$20.0 N$$. (a) Draw a free body diagram of the suitcase. (b) What angle does the strap make with the horizontal? (c) What is the magnitude of the normal force that the ground exerts on the suitcase?

Solution
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(a) See the free-body diagram of the suitcase in figure below.

(b) $$m_{suitcase} = 20.0 kg, F = 35.0 N$$
$$\sum F_{x} = ma_{x} : −20.0 N + F \cosθ = 0$$
$$\sum F_{y} = ma_{y} : +n + F \sinθ − F_{g} = 0$$
$$F \cosθ = 20.0 N$$
$$\cos\theta=\frac{20.0N}{35.0N}=0.571$$
$$\theta=55.2^{0}$$

(c) With $$F_{g} = (20.0 kg)(9.80 m/s^{2})$$,
$$n = F_{g} − F \sinθ = [196 N−(35.0 N)(0.821)]$$
$$n = 167 N$$

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