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Consider the 65.0-kg ice skater being pushed by two others
Chapter 5, Problem 7(choose chapter or problem)
Consider the \(65.0-\mathrm{kg}\) ice skater being pushed by two others shown in Figure \(5.21\).
(a) Find the direction and magnitude of \(\mathbf{F}_{\text {tot }}\) , the total force exerted on her by the others, given that the magnitudes \(F_{1}\) and \(F_{2}\) are \(26.4 \mathrm{~N}\) and \(18.6 \mathrm{~N}\), respectively.
(b) What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of \(\mathbf{F}_{\text {tot }}\) ?
(c) What is her acceleration assuming she is already moving in the direction of \(\mathbf{F}_{\text {tot }}\)? (Remember that friction always acts in the direction opposite that of motion or attempted motion between surfaces in contact.)
Figure 5.21
Questions & Answers
QUESTION:
Consider the \(65.0-\mathrm{kg}\) ice skater being pushed by two others shown in Figure \(5.21\).
(a) Find the direction and magnitude of \(\mathbf{F}_{\text {tot }}\) , the total force exerted on her by the others, given that the magnitudes \(F_{1}\) and \(F_{2}\) are \(26.4 \mathrm{~N}\) and \(18.6 \mathrm{~N}\), respectively.
(b) What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of \(\mathbf{F}_{\text {tot }}\) ?
(c) What is her acceleration assuming she is already moving in the direction of \(\mathbf{F}_{\text {tot }}\)? (Remember that friction always acts in the direction opposite that of motion or attempted motion between surfaces in contact.)
Figure 5.21
ANSWER:Step 1 of 8
Given data:
The mass of the skater m is given as 65 kg.
The forces are given as \({F_1} = 26.4\;{\rm{N,}}{F_2} = 18.6\;{\rm{N}}\).
Step 2 of 8
(a)
The magnitude of the total force \({F_{tot}}\) is:
\({F_{tot}} = \sqrt {F_1^2 + F_2^2}\)
Substitute the values in the above expression, and we get,
\({F_{tot}} = \sqrt {{{26.4}^2} + {{18.6}^2}} \)
\({F_{tot}} = 32.29\;{\rm{N}}\)
Now, the angle made by the total force is:
\(\theta = {\tan ^{ - 1}}\left( {\frac{{{F_2}}}{{{F_1}}}} \right) \)
Substitute the values in the above expression, and we get,
\(\theta