A box with a mass of 50kg is being held in place on a frictionless ramp inclined at 25 degrees to the horizontal. And a 70kg on another ramp at 20 degrees. If both are held at the same position...

A box with a mass of 50kg is being held in place on a frictionless ramp inclined at 25 degrees to the horizontal. And a 70kg on another ramp at 20 degrees. 

If both are held at the same position to cover the same amount of distance, Which one will reach the bottom first? And at what should be the angel for the 2nd ramp so both boxes will reach the bottom at the same time?

MUST be solved using vectors

Asked on by layalom

2 Answers | Add Yours

quantatanu's profile pic

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted on

I am sorry, it must be a technical problem of the page probably, because I had used bold letters for "i", "j", "r1","r2","a1"and "a2" as these are vectors, where "i","j" are unit vectors along x, y directions, but after submitting the bold nature has disappeared!

So I am submitting the same answer again, I am not sure it will work or not:

________________________________________________________

(Bold faced letters will indicate vectors)

Consider the origine to be the bottom point of the inclined ramp, and consider the plane to be x-y plane, so co-ordinates for various quantities are:

Position vector of the box of mass m1 (50 kg) =

 r1 = h1 Cos(25°)  i + h1 Sin(25°)  j

where h1 is the distance of the box of mass m1=50kg, along the ramp from the origin.

Acceleration of box with mass m1 = a1 = 0  i - g  j

Position vector of the box of mass m2 (70 kg) =

 r2= h2 Cos(20°)  i + h2 Sin(20°)  j

where h2 = h1, is the distance of the box of mass m2= 70kg, along the ramp from the origin.

Acceleration of box with mass m1 =a2 = 0  i - g  j


First Part of The Problem:


Notice that both of these boxes have the same acceleration along y-direction,  

a1 = a2 = - g  j,  and hence the box that has smaller y-component of the position vector is actually travelling lesser along the y-direction and will reach earlier, 

Component of r1 along y-direction =r1_y = h1 Cos(25°)

Component of r2 along y-direction = r2_y = h2 Cos(20°) = h1 Cos(20°)

 Clearly 

r1_y < r2_y,       because, Cos(25°) < Cos(20°)

So the box with mass 50 kg will reach the bottom first.

 


Second Part of The Problem:


The angle of the second ramp so that both boxes reaches the bottom exactly at the same time, given  h1=h2, should be the same as that of the first ramp. 

Reason: Acceleration is the same for both so its the distance along y-direction that matters, and as h1=h2 so its the angle that matters, because we need

r1_y = r2_y

and  

r1_y = h1 Cos(25°)

so r2_y must also be Cos(25°)

so angle for the second ramp should also be  25°.

 

quantatanu's profile pic

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted on

(Bold faced letters will indicate vectors)

Consider the origine to be the bottom point of the inclined ramp, and consider the plane to be x-y plane, so co-ordinates for various quantities are:

Position vector of the box of mass m1 (50 kg) =

 r1 = h1 Cos(25°) i + h1 Sin(25°) j

where h1 is the distance of the box of mass m1=50kg, along the ramp from the origin.

Acceleration of box with mass m1 = a1 = 0i - gj

Position vector of the box of mass m2 (70 kg) =

 r2= h2 Cos(20°)i + h2 Sin(20°) j

where h2 = h1, is the distance of the box of mass m2= 70kg, along the ramp from the origin.

Acceleration of box with mass m1 =a2 = 0i - g j


First Part of The Problem:


Notice that both of these boxes have the same acceleration alongy-direction,  

a1 = a2 = - g j, and hence the box that has smaller y-component of the position vector is actually travelling lesser along the y-direction and will reach earlier, 

Component of r1 along y-direction =r1_y = h1 Cos(25°)

Component of r2 along y-direction = r2_y = h2 Cos(20°) = h1 Cos(20°)

 Clearly 

r1_y < r2_y,       because, Cos(25°) < Cos(20°)

So the box with mass 50 kg will reach the bottom first.

 


Second Part of The Problem:


The angle of the second ramp so that both boxes reaches the bottom exactly at the same time, given  h1=h2, should be the same as that of the first ramp. 

Reason: Acceleration is the same for both so its the distance along y-direction that matters, and as h1=h2 so its the angle that matters, because we need

r1_y = r2_y

and  

r1_y = h1 Cos(25°)

so r2_y must also be Cos(25°)

so angle for the second ramp should also be  25°.

 

 

 

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