SOLUTION: A taho vendor carries a 1.5 m long light plank over his shoulder. At the ends of the plank are two buckets weighing 40.0 N and 60.0 N, respectively. (a) Find the value of force F e

Question 1186510: A taho vendor carries a 1.5 m long light plank over his shoulder. At the ends of the plank are two buckets weighing 40.0 N and 60.0 N, respectively. (a) Find the value of force F exerted by his shoulder. Neglect the weight of the plank. (b) Where should he support the plank for it to be balanced horizontally?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
**(a) Force Exerted by the Shoulder**
Since the plank is in equilibrium, the upward force (F) exerted by the shoulder must balance the total downward weight of the two buckets. Therefore:
F = Weight of Bucket 1 + Weight of Bucket 2
F = 40.0 N + 60.0 N
F = 100.0 N
The vendor's shoulder must exert a force of 100.0 N upwards.
**(b) Support Point for Horizontal Balance**
For the plank to be balanced horizontally, the torques (rotational forces) on either side of the support point must be equal. Torque is calculated as the force multiplied by the distance from the pivot point (in this case, the shoulder).
Let's choose the support point to be a distance 'x' from the 40.0 N bucket. This means it will be (1.5 m - x) away from the 60.0 N bucket.
Torque due to Bucket 1 = Torque due to Bucket 2
(40.0 N) * x = (60.0 N) * (1.5 m - x)
Now, solve for x:
40x = 90 - 60x
100x = 90
x = 0.9 m
The vendor should support the plank 0.9 meters away from the 40.0 N bucket.

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