<u>Answer:</u>
The force F applied to the handle = 330.03 N
<u>Explanation:</u>
The force can be resolved in to two, horizontal component and vertical component. If θ is the angle between horizontal and applied force we have
Horizontal component of force = F cos θ
Vertical component of force = F sin θ
In this problem normal force exerted on the suitcase is 160 N, that is vertical component of force = 160 N and angle θ = 29⁰.
So, F sin 29 = 160
F = 330.03 N
The force F applied to the handle = 330.03 N
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