the floor of a lorry is 2.0m high,a plank 5m long is used as an inclined lane to raise some load up the lorry,if the efficiency of this machine is 50%,what is minimum effort applied parallel to plane required to raise 200N load up the plane
To solve this problem, we need to consider the work done by the effort and the work done against gravity.
The work done by the effort is given by W_effort = F_effort * d_effort * cos(theta), where F_effort is the effort force, d_effort is the distance moved by the effort, and theta is the angle between the effort force and the horizontal direction.
The work done against gravity is given by W_gravity = m * g * h, where m is the mass of the load, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the lorry floor.
Since the machine has an efficiency of 50%, the work done by the effort is equal to 50% of the work done against gravity. Therefore, we can write:
0.5 * W_effort = W_gravity
We know that the load is 200 N, so we can substitute the values into the equation:
0.5 * (F_effort * d_effort * cos(theta)) = 200 * 9.8 * 2.0
Simplifying, we get:
F_effort * d_effort * cos(theta) = 200 * 9.8 * 2.0 * 2
Dividing both sides by cos(theta), we get:
F_effort * d_effort = (200 * 9.8 * 2.0 * 2) / cos(theta)
Now we need to find the angle theta. The length of the plank is 5 m, and the height of the lorry floor is 2 m, so we can use the trigonometric formula:
sin(theta) = h / d_plank
Substituting the values, we have:
sin(theta) = 2 / 5
theta = arcsin(2/5)
Now we can substitute the value of theta and solve for F_effort:
F_effort * d_effort = (200 * 9.8 * 2.0 * 2) / cos(arcsin(2/5))
Finally, we divide both sides by d_effort to find the minimum effort applied parallel to the plane:
F_effort = (200 * 9.8 * 2.0 * 2) / (d_effort * cos(arcsin(2/5)))