A) What force T do you need to exert to accomplish this? B) What is the magnitud

Question

A) What force T do you need to exert to accomplish this?

B) What is the magnitude f of the friction force on the upper box?

C)What is the direction of the friction force on the upper box?

lowering two boxes, one on top of the other, down the ramp shown in the figure (Figure 1) by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 13.0cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.431, and the coefficient of static friction between the two boxes is 0.772. A) What force T do you need to exert to accomplish this? B) What is the magnitude f of the friction force on the upper box? C)What is the direction of the friction force on the upper box?

Explanation / Answer

The amount of force needed to complete this task is about 45.166984 newtons of force. Here's how I got my answer...Oh yeah, This problem will require basic trig, so a scientific calculator will come in handy

First, I found the angle of the ramp. This is the angle the ramp is raised to above the horizon. Using basic trigonometry, I found that the 2 boxes were sitting on a plane inclined at about 27.7585406 degrees above the horizon.

Since the two boxes are sitting on top of each other, I just went ahead and added up their masses as one box, 80kg.

After all that, I began to find the forces acting on the box that made it accelerate. First, I found the normal force of the box acting on the ramp by this equation:
cos27.7585406*784.8=694.4838N

With this info, I can find the force of friction.
force of friction=694.4838209(.444)=308.3508N

Now, I begin to work on finding the forces working to pull the box down the ramp, which is in the x-axis.
I did this by using this equation:
sin27.7585406*784.8=365.5178N

With that, I will simply use the equation, f=ma
365.5178005/80=4.568972506m/s^2. This will be the acceleraton in the absence of friction.

When we add friction to this, we get a de-acceleration.
the force of friction causes a de-acceleration of 3.854385206m/s^2, so this means the actual acceleration of the box is .7145873m/s^2

Since there is someone is exerting a force on the system, the speed goes down to .15m/s.
To find the force the person exerts on the system, I found the difference of acceleration, .7145873-.15=.5645873m/s

Finally, I am going to go back the equation, f=ma
The person made a de-acceleration of .5645873m/s^2 on an 80kg box, so

f=.56873*80=45.166984N

as figure is not mentioned so i took some values on my own

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