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

Answer 1

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)))