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# Tutor profile: Davis P.

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Davis P.
Mechanical Engineer
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A man weighing 80kg steps on a scale (reading in Newtons) in an elevator. The elevator accelerates at a rate of 3m/\$\$s^2\$\$ before reaching a constant speed between floors. What does the scale read between floors? If the man is traveling up, what does the scale read just before the elevator stops?

Inactive
Davis P.

The first step to approaching this problem is to draw a free body diagram. Because the elevator is traveling a constant speed between floor, no acceleration is present and net force on the man is zero. In this circumstance the scale reads the man's weight accurately as 800N. We know this from summing the forces in the Y direction and the resulting equation: \$\$F_{scale}\$\$ = W where W is the man's weight in N. Since W = 80kg * 10m/\$\$s^2\$\$ = 800N, then the scale will read 800N. (The acceleration de to gravity is approximated as 10m/\$\$s^2\$\$ Just before the elevator comes to a stop, the net force is not equal to 0. Summing the force in the Y direction, we get: 80kg*(-3m/\$\$s^2\$\$) = \$\$F_{scale}\$\$ - W or 80kg*(-3m/\$\$s^2\$\$) = \$\$F_{scale}\$\$ - 800N Solving for \$\$F_{scale}\$\$, we get \$\$F_{scale}\$\$ = 80kg*(-3m/\$\$s^2\$\$) + 800N or \$\$F_{scale}\$\$ = 560N Thus, the scale is reading 560N as it is coming to a stop on the man's way up to his floor.

### Subject:Trigonometry

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Question:

The area of a right triangle is 60. One of it's angles is 27 degrees. Find the lengths of the sides (a and b) and hypotenuse (c) of the triangle.

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Davis P.

Solve for the sides using a system of equations: [Eq 1] Area = ab = 60 [Eq 2] b/a = tan(27) Solving Eq 2 for B yields: b = a*tan(27) Substitute for b in Eq 1, then solve for a: (\$\$a^2\$\$)*tan(27) = 60 a \$\$\approx\$\$ 10.8 --- (the negative solution to the square being thrown out) Thus from Eq 1, b \$\$\approx\$\$ 5.5 From Pythagorean Theorem: \$\$a^2\$\$ + \$\$b^2\$\$ = \$\$c^2\$\$ \$\$10.8^2\$\$ + \$\$5.5^2\$\$ \$\$\approx\$\$ \$\$c^2\$\$ c \$\$\approx\$\$ 12.1

### Subject:Algebra

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Question:

10x-3y=7 2x-y=3 Solve for x and y

Inactive
Davis P.

Solve the second equation for y: y=2x-3 Substitute for y in the first equation: 10x-3(2x-3)=7 10x-6x+9= 7 4x=-2 x=-1/2 Substitute value for x into either equation: 10(-1/2) - 3y = 7 y=-4

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