A force of 35lb is required to hold a 72lb toolbox on an incline. What angle does the incline make with the horizontal?
Please help me set this up.
wt. of toolbx acts vertically downwards and force acts along the inclined plane.
resolve the wt.into 72cos¡è¢ã 72sin
wt. of toolbx acts vertically downwards and force acts along the inclined plane.
resolve the wt.into 72cos¡è ¡ý, 72sin¡è ¡û
apply conditions of equilibrium,
wt. of toolbx acts vertically downwards and force acts along the inclined plane.
resolve the wt.into 72cos¡è ¡ý, 72sin¡è ¡û
apply conditions of equilibrium, ie Fx=0
72sin¡è=35
then solve by taking inverse of sin¡è
¡è=29.!
wt. of toolbx acts vertically downwards and force acts along the inclined plane.
resolve the wt.into 72cos¡è ¡ý, 72sin¡è ¡û
apply conditions of equilibrium, ie Fx=0
72sin¡è=35
then solve by taking inverse of sin¡è
¡è=29.09.
thank you
To set up this problem, we can use the concept of equilibrium. When an object is in equilibrium, the sum of the forces acting on it is zero. In this case, the force required to hold the toolbox on the incline is equal to the component of the weight of the toolbox parallel to the incline.
Let's start by drawing a diagram to visualize the situation. Draw a horizontal line to represent the incline, and label it as the horizontal direction. Then draw a line perpendicular to the incline to represent the vertical direction.
Label the force required to hold the toolbox (35 lb) and the weight of the toolbox (72 lb). The weight of the toolbox can be split into two components: one perpendicular to the incline and one parallel to the incline.
Now, let's analyze the forces acting on the toolbox.
1. The weight of the toolbox can be split into two components:
- The component perpendicular to the incline (mgcosθ) does not affect the force required to hold the toolbox on the incline.
- The component parallel to the incline (mgsinθ) contributes to the force required to hold the toolbox.
2. The force required to hold the toolbox on the incline (35 lb) acts parallel to the incline.
Since the toolbox is in equilibrium, the sum of the forces acting on it is zero in the horizontal and vertical directions.
Equating the forces in the horizontal direction:
Force required to hold toolbox = mgsinθ
35 lb = 72 lb * sinθ
Solving for θ:
sinθ = 35 lb / 72 lb
θ = arcsin(35/72)
Using a calculator, you can find the inverse sine (arcsine) of 35/72 to get the value of θ.