Dynamics - Relative Motion Help?

[Khamul]

Homework Statement

Conveyor belt A, which forms a 20° angle with the horizontal, moves at a constant speed of 4 ft/s and is used to load an airplane. Knowing that a worker tosses duffel bag B with an initial velocity of 2.5 ft/s at an angle of 30° with the horizontal, determine the velocity of the bag relative to the belt as it lands on the belt.

Homework Equations

V_i,x = V_i*cos($$\theta$$)
V_i,y = V_i*sin($$\theta$$)
(x_f-x_i) =V_i,x*t
(y_f-y_i) =V_i,y*t - (1/2)*g*t²
r_B = r_A+r_B/A
v_B = v_A+v_B/A
a_B = a_A+a_B/A

The Attempt at a Solution

I think I've gotten the majority of this problem tackled, so I will list what I have done (as I am highly likely to have done something wrong.) As you can see from the picture below, a linear equation (I think?) is required out of the conveyor belt, so that is the first thing I did.

We know that to make a right triangle with the belt, one side is 90, we're given 20, so the final angle is 70. For pure relations' sake to find a slope, I set the bottom length equal to 10, used tan(20)=opp/10, and found the rise/run to be 3.64/10 which came out to be a slope of .364.

I also set my origin at the location of the duffel bag, which was 1.5 feet above the starting location of the conveyor belt line. So, for my linear equation, I found y=.364x-1.5

Then from there, (I think this is right?) I subbed my x_f and y_f equations into my linear equation. I was then faced with a quadratic, which, when solved for t, I found t=.3209 s. Fast forwarding a bit more after this I found my final x and y distances.

My x final distance was .6947 ft, while my y final distance was -1.2465 ft.

And...this is where I taper off. I am unsure how to proceed next. I know my final distances, the time it took to hit the conveyor belt...but I am not sure how to relate all of this to find the relative velocity of the bag to the belt.

Could anyone spare some help? I would greatly appreciate it!

4. F.B.D.

 

[tiny-tim]

Hi Khamul!

My x final distance was .6947 ft, while my y final distance was -1.2465 ft.

And...this is where I taper off. I am unsure how to proceed next. I know my final distances, the time it took to hit the conveyor belt...but I am not sure how to relate all of this to find the relative velocity of the bag to the belt.

now you need to find the x and y components of the final velocity

to find the relative velocity, you then just subtract the velocity components of the belt

btw, why did you do all this, instead of simply saying tan20° = 0.364 ? …

We know that to make a right triangle with the belt, one side is 90, we're given 20, so the final angle is 70. For pure relations' sake to find a slope, I set the bottom length equal to 10, used tan(20)=opp/10, and found the rise/run to be 3.64/10 which came out to be a slope of .364.

 

[Khamul]

Hi Khamul!


now you need to find the x and y components of the final velocity

to find the relative velocity, you then just subtract the velocity components of the belt

btw, why did you do all this, instead of simply saying tan20° = 0.364 ? …

Goood morning tiny-tim, thanks for getting around to helping me so early!

Huh. Well hey, look at that! I learned something new about slopes, thank you! Also, thank you for steering me in the right direction I was burned out between this and thermodynamics by midnight last night!

 

[waterwoman27]

I am actually working on this problem right now and my question for you is where did you place you axis in this problem?

 

[DeeNos]

of the duffel bag:To find the relative velocity of the bag to the belt, we need to consider the velocities of both the bag and the belt. The bag's velocity includes both its initial velocity and the velocity it gains from being thrown at an angle. The belt's velocity is constant at 4 ft/s.

Using the equations listed in the problem, we can find the x and y components of the bag's initial velocity:

Vi,x = 2.5 ft/s * cos(30°) = 2.165 ft/s
Vi,y = 2.5 ft/s * sin(30°) = 1.25 ft/s

To find the bag's final velocity, we can use the equation vB = vA + vB/A, where vA is the velocity of the belt and vB/A is the velocity of the bag relative to the belt. We know that vA = 4 ft/s, so we can plug in the values we found for the x and y components of the bag's initial velocity:

vB = 4 ft/s + 2.165 ft/s * i + 1.25 ft/s * j

where i and j are unit vectors in the x and y directions, respectively.

To find the magnitude and direction of the bag's final velocity, we can use the Pythagorean theorem and the inverse tangent function:

|vB| = sqrt((4 ft/s)^2 + (2.165 ft/s)^2 + (1.25 ft/s)^2) = 4.68 ft/s
θ = tan^-1(1.25 ft/s / 2.165 ft/s) = 30.2°

Therefore, the final velocity of the bag relative to the belt is 4.68 ft/s at an angle of 30.2° above the horizontal.