A rock of mass 170 kg needs to be lifted off the ground. One end of a metal bar is slipped under the rock, and a fulcrum is set
up under the bar at a point that is 0.65 m from the rock. A worker pushes down (perpendicular) on the other end of the bar, which is 1.9 m away from the fulcrum. What force is required to move the rock?
1 answer:
Answer:
866.32 N
Explanation:
The diagram explains better.
Taking the total moment of forces, the sum of the clockwise moment of forces about the fulcrum must be equal to the sum of the anticlockwise moment of forces about the fulcrum:
F * (1.9 - 0.65) = 170 * 9.8 * 0.65
F * 1.25 = 1082.9
F = 1082.9/1.25
F = 866.32 N
A force of 866.32 N is needed to move the rock.
You might be interested in
Answer:
Mass of the climber = 69.38 kg
Explanation:
Change in length
Load, P = m x 9.81 = 9.81m
Young's modulus, Y = 0.37 x 10¹⁰ N/m²
Area
Length, L = 15 m
ΔL = 5.1 cm = 0.051 m
Substituting
Mass of the climber = 69.38 kg
a = 3.09 m/s²
<h3>Explanation</h3>This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the <em>timeless </em>suvat equation is likely what you need for this question.
In the <em>timeless</em> suvat equation,
where
is the acceleration of the car;
is the <em>final</em> velocity of the car;
is the <em>initial</em> velocity of the car; and
is the displacement of the car.
Note that <em>v</em> and <em>u</em> are velocities. Make sure that you include their signs in the calculation.
In this question,
;
; and
.
Apply the <em>timeless</em> suvat equation:
.
The value of is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.