A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle θ above the horizontal (see figure). She pulls on the strap with a 35.0-N force, and the friction force on the suitcase is 20.0 N.?

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greishava9g6 · Accepted Answer

She is moving at a constant speed, so neither her force nor the friction will cause her to accelerate or decelerate. As a result, we can say that the friction and the horizontal component of her 35.0 N are equal but opposite.      20.0    N    =    35.0    N    ⋅          cos      θ      Solving for             cos      θ      ,            cos      θ        =                  20.0        N                    35.0        N              =    0.5714        θ    =          arccos              (        0.5714        )              =          55.2      ∘      The weight of the suitcase is             weight        =    m    ⋅    g    =    20.0    k    g    ⋅    9.8          m              s        2              =    196    N   However, it is not the response to the query. The suitcase is being partially raised by the vertical component of her 35.0 N. The partial lift&#039;s value is      35.0    N    ⋅                            sin          55.2                ∘            =        28.7    N  That leaves       196    N    -    28.7    N    =    167.3    N   that the normal force needs to provide to maintain vertical equilibrium.

A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle θ above the horizontal (see figure). She pulls on the strap with a 35.0-N force, and the friction force on the suitcase is 20.0 N.?

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