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3 years ago

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"A student bikes to school by traveling first dN = 0.900" and "Similarly, let d⃗ W be the displacement vector corresponding to t

he second leg of the student's trip. Express d⃗ W in component form. Express your answer as two numbers separated by a comma. Be careful with your signs"

1 answer:

raketka [301]3 years ago

6 0

Complete Question

A student bikes to school by traveling first dN = 0.900 miles north, then dW = 0.300 miles west, and finally dS = 0.100 miles south.

Similarly, let d⃗ W be the displacement vector corresponding to the second leg of the student's trip. Express d⃗ W in component form.

Express your answer as two numbers separated by a comma. Be careful with your signs.

Answer:

The value is $dT = ( -0.3, 0.8)$

Explanation:

From the question we are told that

The first displacement is $dN = 0.900 \ due \ North$ i.e positive y-axis

The second displacement is $dW = 0.300 \ miles \ due \ west$ i.e negative x-axis

The final displacement is $dS = 0.100 \ miles \ due \ south$ i.e negative y-axis

Generally dW in component for is

$dW = (-0.3 , 0)$

Generally the total displacement of the student is mathematically represented as

$dT = ( -0.3, (0.90 - 0.10))$

$dT = ( -0.3, 0.8)$

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Answer:

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In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

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we add $\mu_{k}mg$ to both sides so we get:

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Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

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when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

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Nadya [2.5K]

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F = 707 N acting along the plane

Fn = F sin θ + M g cos θ forces acting perpendicular to plane

Fn = 707 * 1/2 + 80 * 9.8 * .866 = 1030 Newtons forces normal to plane

(this would give a coefficient of friction of 100 / 1030 = .097 = Fn)

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