Suppose that a meter stick is balanced at its center. A 0.27-kg is positioned at 18 cm from the left end of the meter stick. Where should a 0.28 kg mass be placed to balance the 0.27 kg mass? Express your answer in terms of the position (in cm) of the 0.28-kg mass as measured from the left end of the meter stick.
Compute the total torque about the center due to the two forces. Mass m1 is 50-18 = 32 cm from the center. If x is the distance of mass m2 from the left end, then it is x-50 to the right of the center pivot. Set the total torque equal to zero and solve for x.
-0.27g*(50-18) + 0.28g(x-50) = 0
You can factor out the g.
0.27*32 = 0.28*(x-50)
Now, we will solve for x:
0.27*32 = 0.28*(x-50)
8.64 = 0.28x - 14
8.64 + 14 = 0.28x
22.64 = 0.28x
Divide both sides by 0.28:
x ≈ 80.86 cm
So the 0.28 kg mass should be placed approximately 80.86 cm from the left end of the meter stick to balance the 0.27 kg mass.
To solve for x, let's first simplify the equation:
0.27 * 32 = 0.28 * (x - 50)
Now, let's perform the calculations:
8.64 = 0.28x - 14
Next, let's isolate the variable x by bringing the constant term to the right side:
0.28x = 8.64 + 14
0.28x = 22.64
Finally, divide both sides of the equation by 0.28 to solve for x:
x = 22.64 / 0.28
x ≈ 80.86 cm
Therefore, the 0.28 kg mass should be placed approximately 80.86 cm from the left end of the meter stick in order to balance the 0.27 kg mass.
To solve the problem, you need to compute the total torque about the center due to the two masses.
First, calculate the torque due to the 0.27 kg mass. The mass is positioned 18 cm from the left end of the meter stick. Since the center is the pivot point, the distance from the center to this mass is 50 cm - 18 cm = 32 cm.
The torque due to the 0.27 kg mass is given by:
Torque1 = 0.27 kg * g * (50 cm - 18 cm)
Next, consider the 0.28 kg mass that needs to be placed to balance the system. Let x be the distance of this mass from the left end of the meter stick. It is x cm to the right of the center pivot.
The torque due to the 0.28 kg mass is given by:
Torque2 = 0.28 kg * g * (x cm - 50 cm)
The total torque about the center is the sum of Torque1 and Torque2. Since the system is balanced, the total torque should be zero.
Setting the total torque equal to zero, we have:
0.27 kg * g * (50 cm - 18 cm) + 0.28 kg * g * (x cm - 50 cm) = 0
You can now solve this equation for x.
You can factor out g from both terms on the left-hand side of the equation:
0.27 * 32 cm = 0.28 * (x cm - 50 cm)
Simplifying the equation, we get:
8.64 cm = 0.28 * (x cm - 50 cm)
Now, solve for x by dividing both sides of the equation by 0.28:
8.64 cm / 0.28 = x cm - 50 cm
Finally, add 50 cm to both sides of the equation to isolate x:
x cm = 8.64 cm / 0.28 + 50 cm
Evaluate the expression on the right-hand side to find the position of the 0.28 kg mass as measured from the left end of the meter stick.