Enable contrast version

# Tutor profile: Brian K.

Inactive
Brian K.
Mechanical Engineering student, in the top 10% of my class
Tutor Satisfaction Guarantee

## Questions

### Subject:Mechanical Engineering

TutorMe
Question:

Which of the following is NOT a requirement for equilibrium of a two force member? (more than one answer can be chosen) a) The forces are equal magnitude b) The forces have the same line of action c) The member is straight d) The member has forces applied at two points

Inactive
Brian K.

Going over each answer choice, we will see which is the NOT a requirement a: No A two force member MUST have equal magnitude, or else it will not be in equilibrium b: No A two force member MUST have the same line of action, or else this will result in a net moment c: Yes This is not a requirement, since a two force member can have any shape as long as there are two forces that share a line of action and have the same magnitude d: No This is a requirement, since if you have more than two forces, it is not considered a two force member

### Subject:Calculus

TutorMe
Question:

Find the rate at which G(x,y,z) = 3e^(2x-y) + tan((pi)z/4) is changing at the point (1,2,1) in the direction towards the point (2,3,2)

Inactive
Brian K.

**I will be using partial derivatives, so when you see any derivative notation, such as df/dx, I'm assuming partial derivative, unless otherwise stated** We know that the directional derivative is equal to $$< dG/dx, dG/dy, dG/dz> * u$$ The asterisk is the dot product (not to be confused with the cross product), and the u is the unit direction vector (usually the unit direction vector has a hat notation on top of it). Now we start to plug in values and solve and we get $$dG/dx= 6e^(2x-y)$$ $$dG/dy= -3e^(2x-y)$$ $$dG/dz= sec^2(\pi z/4) * (\pi/4)$$ Now, we evaluate these answers at the point (1,2,1), and we get dG/dx = 6 dG/dy= -3 $$dG/dz= \pi/2\sqrt(2)$$ The unit direction vector, u, is the direction (2,3,2) minus the point (1,2,1) $$u= <1,1,1>/\sqrt(3)$$ Now, we plug back into the equation of the directional derivative $$< dG/dx, dG/dy, dG/dz> * u$$ and get $$<6,-3,\pi/2\sqrt(2)> * <1,1,1>/\sqrt(3)$$ which becomes $$(3/\sqrt3) + (\pi/2\sqrt2)$$

### Subject:Physics

TutorMe
Question:

A box, which has a mass, m, is on an incline that is raised $$\theta$$. The box is connected to a cord that runs over a pulley, and then attaches to a spring that has a spring constant, k. The height of the incline is h, and the length of the box from the top on the incline is L. When the box is released, the box goes down the incline and then rebounds back up. Assuming that the pulley has no mass and is frictionless, and the coefficient of kinetic friction is $$\mu_k$$, what is the acceleration, a, at any L before the rebound? (in other words, what is acceleration before a rebound such that L < Lmax)

Inactive
Brian K.

From this scenario, we want to draw the free body diagram. In this diagram, we know the acceleration is downwards, so that is where we can choose our positive and negative values. Any forces that go in the opposite direction of the acceleration is negative, and any forces that go with the acceleration is positive. We know that the force that goes with the acceleration is a component of the weight, $$mgsin(\theta)$$. the forces that go against the acceleration are the spring force, S, and the friction force, $$f_k$$. Therefore, we know the equation is $$ma= mgsin(\theta) - S - f_k$$ We know that the spring force is kL and the friction force is $$\mu_k N$$, where N is the normal force. The normal force is also equal to $$mgcos(\theta)$$. Therefore, we get $$ma= mgsin(\theta) - kL - \mu_k mgcos(\theta)$$ solving for a, we get $$a= g[sin(\theta) - \mu_k cos(\theta)] - (kL/m)$$

## Contact tutor

Send a message explaining your
needs and Brian will reply soon.
Contact Brian