A body of mass kg suspended at the 15cm mark of a ruler balanced if at the 30cm mark on a pivot.if the ruler has a mass of 36g and it entime a gravity is at 55cm mark.culate (I)mass (ii) distance of the balance point of the ruler from the 3 and when the body is removed to the 25cm mark.
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projectile
To solve this problem, we need to use the concept of torques, which is the rotational equivalent of force. The formula for torque is given by:
Torque = force × distance from the pivot
First, let's calculate the torque exerted by the ruler on the pivot when the body is suspended at the 15cm mark:
Torque (ruler) = Mass (ruler) × Gravity (ruler) × Distance (ruler from pivot)
= 0.036 kg × 9.8 m/s^2 × 0.55 m
= 0.1926 Nm
Since the ruler is balanced, the torque exerted by the body also needs to be equal and opposite to the torque exerted by the ruler:
Torque (body) = Mass (body) × Gravity (body) × Distance (body from pivot)
= Mass (body) × 9.8 m/s^2 × 0.15 m
Now we can solve for the mass of the body (i) using the equation above. Since we don't know the value of gravity of the body, we'll need to use some algebra to solve for the mass. Additionally, the units are given in grams, so we'll need to convert them to kilograms.
0.036 kg × 9.8 m/s^2 × 0.55 m = Mass (body) × Gravity (body) × 0.15 m
Simplifying the equation, we find:
0.036 kg × 9.8 m/s^2 × 0.55 m = Mass (body) × Gravity (body) × 0.15 m
0.1926 Nm = Mass (body) × Gravity (body) × 0.15 m
Dividing both sides of the equation by 0.15 m, we get:
Mass (body) × Gravity (body) = 0.1926 Nm / 0.15 m
Mass (body) = (0.1926 Nm / 0.15 m) / Gravity (body)
Since the gravitational acceleration is always 9.8 m/s^2, we know that Gravity (body) = 9.8 m/s^2. Plugging these values into the equation, we can calculate the mass (i) of the body.
For (ii), we need to find the new balance point of the ruler when the body is moved to the 25cm mark. Here's how we can calculate it:
First, let's calculate the new torque exerted by the ruler on the pivot:
Torque (ruler-new) = Mass (ruler) × Gravity (ruler) × Distance (ruler from pivot)
= 0.036 kg × 9.8 m/s^2 × 0.25 m
Since the torque exerted by the body must be equal and opposite to the torque exerted by the ruler:
Torque (body-new) = Mass (body) × Gravity (body) × Distance (body from pivot)
= Mass (body) × 9.8 m/s^2 × 0.25 m
Now we can set the torques equal to each other and solve for the new balance point:
Mass (ruler) × Gravity (ruler) × Distance (ruler from pivot) = Mass (body) × Gravity (body) × Distance (body from pivot)
0.036 kg × 9.8 m/s^2 × 0.25 m = Mass (body) × 9.8 m/s^2 × Distance (body from pivot)
0.0294 Nm = Mass (body) × 9.8 m/s^2 × Distance (body from pivot)
Dividing both sides by (Mass (body) × 9.8 m/s^2), we find:
Distance (body from pivot) = 0.0294 Nm / (Mass (body) × 9.8 m/s^2)
Now, we can substitute the previously calculated mass (i) into this equation and solve for the distance (ii) of the balance point of the ruler from the pivot when the body is at the 25cm mark.