A box is lifted with a pry bar, by slipping the bar under the box and lifting up on the bar. If the pry bar is 2m in length and it is grasped at the end, and the box is located at the opposite end, the fulcrum, with its center of gravity at a distance of 0.5 m from the fulcrum.
what is the size of the force required to lift the box if the box has a weight of 100N?
CAN SOMEONE TELL ME IF I AM CORRECT?
.5 X 100 = 50Nm / 2m = 25N
0.5m * 100N = F * 2m
Solve for unknown F
F = 25 N
Right answer; bad notation.
Thanks Mr. drwls for your input, but I do not follow the above notation. is * multiplication? forgive my ignorance...:(
Yes * means multiply
To find the force required to lift the box using a pry bar, we can use the principle of moments. The principle of moments states that the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
In this case, the weight of the box creates a clockwise moment, while the force applied to lift the box creates an anticlockwise moment.
First, let's calculate the clockwise moment caused by the weight of the box. The moment is equal to the weight multiplied by the perpendicular distance between the weight and the fulcrum. In this case, the distance is 0.5m.
Clockwise moment = 100N * 0.5m = 50Nm
Next, let's calculate the anticlockwise moment created by the force applied to the pry bar. The moment is equal to the force applied multiplied by the perpendicular distance between the force and the fulcrum. In this case, the distance is 2m.
Anticlockwise moment = Force * 2m
According to the principle of moments, the clockwise moment is equal to the anticlockwise moment. Therefore:
50Nm = Force * 2m
Solving for Force:
Force = 50Nm / 2m
Force = 25N
So, the force required to lift the box is indeed 25N. Your calculation is correct!