A balance scale has unequal arms. The scale is balanced with a m = 1.2-kg block on the left pan and a M = 1.91 kg block on the right pan (see figure). If the 1.91-kg block is removed from the right pan and the 1.2-kg block is then moved to the right pan, what mass on the left pan will balance the scale?
Usually you TRY all questions, provide what you think are answers... then when a tutor in your subject area comes on line they will CHECK your answers...
I randomly picked this question in your series of questions...
I am not a physics major (rather a Calculus major) and it appears to me that this is more of a logic problem.
They don't tell you the condition of the balance once the removal of the 1.91 mass is gone, and again when the 1.2 mass is removed. Thus there could be more than one answer.
The most simplified answer is that on the side with the 1.2 is a .71 kg mass (since 1.2+.71 makes up the 1.91 mass), as well there could be a .71 mass on the 1.91 kg side. Thus when you remove the 1.2 and the 1.91 kg masses the balance is still balanced with 0.71 kg on both sides...
But do you see how this opens up the answers to the question by the wording of the question?
Let me echo the comment by Ms Pi. This looks like a homework dump to me. You show no effort and o insight at all. I doubt anyone will do all of your work for you.
I am not a physics major either. Unfortunately we can't see the diagrams nor can we draw them.
The left mass is M1 and the right mass is M2. The left beam is X1 length and the right beam is X2 length so
M1X1 = M2X2
1.2(X1) = 1.9(X2)
(X1/X2) = 1.9/1.2 = 1.58
If we now place the 1.2 kg mass on the RIGHT pan (M2), then
M1X1 = M2X2
M1 = M2(X2/X1)
M1 = 1.2(1/1.58)
M1 = 0.76
Proof: The arm length ratio must stay the same.
0.76*X1 = 1.2X2 and
(X1/X2) = 1.2/0.76 = 1.58
To find the mass on the left pan that will balance the scale, we need to consider the principle of moments.
The principle of moments states that the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point.
In this case, the point we are considering is the fulcrum or the pivot point of the balance scale. Let's denote the distance of the fulcrum from the left pan as "d1" and the distance of the fulcrum from the right pan as "d2".
When the scale is balanced with the 1.2-kg block on the left pan and the 1.91-kg block on the right pan, the moments on both sides are equal. Therefore, we have:
Moment on the left pan = Moment on the right pan
Considering the masses and distances involved, we can write the equation as:
(1.2 kg) x d1 = (1.91 kg) x d2
Now, let's consider the new scenario where the 1.91-kg block is removed from the right pan and the 1.2-kg block is moved to the right pan. In this case, the moments on both sides will still be equal:
Moment on the left pan = Moment on the right pan
The moment on the left pan can be calculated as:
Mass on left pan x d1
The moment on the right pan can be calculated as:
(1.2 kg) x (d2 - d1)
Setting the two moments equal to each other, we have:
Mass on left pan x d1 = (1.2 kg) x (d2 - d1)
Now we can rearrange the equation to solve for the mass on the left pan:
Mass on left pan = (1.2 kg) x (d2 - d1) / d1
Substituting the given values for m = 1.2 kg and M = 1.91 kg, as well as the values for d1 and d2 (which can be measured from the figure or given separately), we can calculate the mass on the left pan that will balance the scale.