Math

Question* 61. Tll the drawing shows box 1 resting on a tabl with box 2 resting on top of box 1. A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2 , and the other end is connected to box 3. The weights of the three boxes are W1=55 N,W2=35 NW_{1}=55 \mathrm{~N}, W_{2}=35 \mathrm{~N}, and W3=28 NW_{3}=28 \mathrm{~N}. Determine the magnitude of the normal force that the table exerts on box 1 .

Studdy Solution

STEP 1

What is this asking? We've got three boxes connected by a rope and pulley, and we need to find the force of the table pushing up on the bottom box. Watch out! Don't forget that the weight of *both* boxes 1 and 2 contribute to the normal force from the table!

STEP 2

1. Analyze the forces on box 2
2. Analyze the forces on the system of boxes 1 and 2

STEP 3

Box 2 has three forces acting on it: its weight W2W_2 pulling downwards, the tension TT in the rope pulling upwards, and the normal force N12N_{12} from box 1 pushing upwards.

STEP 4

Since box 2 isn't accelerating vertically, the net force on it must be zero.
This gives us: T+N12W2=0T + N_{12} - W_2 = 0

STEP 5

We know W2=35W_2 = 35 N and W3=28W_3 = 28 N.
Since box 3 is lighter than box 2, box 3 will accelerate upwards and box 2 will accelerate downwards.
The tension in the rope will be somewhere between the weights of the two boxes.
Let's find the tension.
The net force on box 3 is TW3T - W_3, and the net force on box 2 is W2TW_2 - T.
Since the boxes are connected by a rope, their accelerations have the same magnitude but opposite directions.
Let aa be the magnitude of the acceleration.
Then, using Newton's second law, TW3=m3aT - W_3 = m_3 a and W2T=m2aW_2 - T = m_2 a.
Since W=mgW = mg, we can write m=Wgm = \frac{W}{g}.
So, TW3=W3gaT - W_3 = \frac{W_3}{g} a and W2T=W2gaW_2 - T = \frac{W_2}{g} a.
Adding these two equations, we get W2W3=W2+W3gaW_2 - W_3 = \frac{W_2 + W_3}{g} a, so a=W2W3W2+W3ga = \frac{W_2 - W_3}{W_2 + W_3}g.
Substituting this back into TW3=W3gaT - W_3 = \frac{W_3}{g} a, we get T=W3+W3(W2W3)W2+W3=2W2W3W2+W3=2352835+28=196063=31.11T = W_3 + \frac{W_3(W_2 - W_3)}{W_2 + W_3} = \frac{2W_2 W_3}{W_2 + W_3} = \frac{2 \cdot 35 \cdot 28}{35 + 28} = \frac{1960}{63} = 31.11 N.
Now we can find N12N_{12}: N12=W2T=3531.11=3.89 NN_{12} = W_2 - T = 35 - 31.11 = \textbf{3.89 N}.

STEP 6

The table exerts a normal force NN upwards on box 1.
The weight of box 1 (W1W_1) and the weight of box 2 (W2W_2) act downwards.
The tension TT also pulls downwards on box 2.

STEP 7

Since boxes 1 and 2 are not accelerating vertically, the net force on the combined system must be zero.
This gives us: NW1W2+T=0N - W_1 - W_2 + T = 0 Notice that the normal force between box 1 and box 2 does *not* appear here because it's an internal force within the system of boxes 1 and 2.

STEP 8

We have W1=55W_1 = 55 N, W2=35W_2 = 35 N, and we found T=31.11T = 31.11 N.
Therefore, N=W1+W2T=55+3531.11=58.89 NN = W_1 + W_2 - T = 55 + 35 - 31.11 = \textbf{58.89 N}.

STEP 9

The magnitude of the normal force that the table exerts on box 1 is **58.89 N**.

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