A worker sits at one end of a 183-N uniform rod that is 2.80 m long. A weight of 107 N is placed at the other end of the rod. The rod is balanced when the pivot is 0.670 m from the worker. Calculate the weight of the worker
Well, it sounds like we're dealing with a balanced rod and a worker who's pulling a tricky balancing act. Let's try to find the weight of the worker, shall we?
To keep things balanced, the total torque on the rod must be zero. Torque is calculated by multiplying the force applied to an object by the distance from the pivot point.
The torque due to the worker can be calculated as follows:
Torque(worker) = force(worker) × distance(worker)
And the torque due to the weight can be calculated as follows:
Torque(weight) = force(weight) × distance(weight)
Since the rod is balanced, we can set these two torques equal to each other and solve for the force of the worker:
force(worker) × distance(worker) = force(weight) × distance(weight)
And if we plug in the given values:
force(worker) × 0.670 m = 107 N × 2.80 m
Now we can solve for the force of the worker:
force(worker) = (107 N × 2.80 m) / 0.670 m
And, drumroll please... we find that the force of the worker is approximately 448 N.
So, the weight of the worker is about 448 N. Remember, this calculation assumes the worker is weightless, but we all know that's just clowning around!
To calculate the weight of the worker, we can use the condition of rotational equilibrium, which states that the sum of the clockwise torques must be equal to the sum of the counterclockwise torques.
Let's denote the weight of the worker as W_w, the weight at the other end of the rod as W_o, the distance from the pivot to the worker as d_w, and the distance from the pivot to the other weight as d_o.
The torque exerted by a force is given by the equation:
Torque = Force * Distance
In this case, the worker's weight creates a counterclockwise torque, while the weight at the other end creates a clockwise torque. The equation for rotational equilibrium can be written as:
W_w * d_w = W_o * d_o
Given:
W_o = 107 N
d_o = 2.8 m
d_w = 0.670 m
Substituting these values into the equation, we get:
W_w * 0.670 m = 107 N * 2.8 m
Now, we can solve for W_w:
W_w = (107 N * 2.8 m) / 0.670 m
Calculating this equation, we find:
W_w ≈ 447.16 N
Therefore, the weight of the worker is approximately 447.16 N.
To calculate the weight of the worker, we can use the principle of torque equilibrium. Torque is a measure of the rotational force experienced by an object. In order for the rod to be balanced, the torques on both ends of the rod must be equal.
Torque is given by the equation:
Torque = Force × Distance
At one end of the rod, the worker exerts a force (weight) on the rod. Let's call this force Fw. The distance from the pivot point to the worker is 0.670 m. So the torque exerted by the worker is:
Torque worker = Fw × 0.670
At the other end of the rod, a weight of 107 N is placed. The distance from the pivot to this weight is 2.80 m. So the torque exerted by this weight is:
Torque weight = 107 × 2.80
Since the rod is balanced, the torques exerted by the worker and the weight should be equal. Therefore, we can set up the following equation:
Fw × 0.670 = 107 × 2.80
To calculate the weight of the worker (Fw), we can rearrange the equation:
Fw = (107 × 2.80) / 0.670
Now plugging in the values and calculating:
Fw = 297.6 / 0.670
Fw ≈ 444.18 N
Therefore, the weight of the worker is approximately 444.18 N.
W•0.67 = 183•(1.4-0.67) +107•(2.8-0.67)
W={183•0.73+107•2.13}/0.67=540 N