A body of mass 7.5kg is to be pulled up along a plane which is inclined at 30degree to the horizontal. If the efficiency of the plane is 75%,what is the minimum force required to pull the body up the plane?
gravitational force parallel to the plane is...m * g * sin(30º)
If the efficiency of the plane is 75%, then the coefficient of friction is .25
the normal force is...m * g * cos(30º)
so the minimum force is
[m * g * sin(30º)] +
[m * g * cos(30º) * .25]
To find the minimum force required to pull the body up the inclined plane, we need to consider the forces acting on the body.
First, let's determine the weight of the body. The weight of an object is given by the formula:
Weight = mass * acceleration due to gravity
Weight = 7.5 kg * 9.8 m/s² (acceleration due to gravity ≈ 9.8 m/s²)
Weight = 73.5 N
Next, we need to find the force required to overcome the weight along the inclined plane. This force is given by:
Force = Weight * sine(incline angle)
In this case, the incline angle is 30 degrees. So, the force required is:
Force = 73.5 N * sine(30°)
Force = 37.75 N
Now, we have the force required to overcome the weight. However, we need to account for the efficiency of the inclined plane, which is given as 75%.
Efficiency = (Useful output / Total input) * 100
The useful output is the force required to overcome the weight, and the total input is the force we need to find.
So, we can rearrange the formula to solve for the force:
Force = (Efficiency / 100) * (Weight * sine(incline angle))
Plugging in the values, we get:
Force = (75 / 100) * (73.5 N * sine(30°))
Force = 27.81 N
Therefore, the minimum force required to pull the body up the inclined plane is approximately 27.81 Newtons.