I need help setting this problem up. A force of 36lb is required to hold a 70lb toolbox on an incline. What angle does the incline make with the horizontal?
need to know mu, friction coef
if there is no friction, let mu = 0
F - mu (70) cos A = (70) sin A
if mu is zero then
F = 90 sin A
so
sin A = 36/90
I got an answer of 23 degree. Is that what you got?
p.s thank you for helping me.
To set up this problem, we can use the concept of forces and components. Here's how we can do it:
First, let's draw a diagram to visualize the situation. We have an incline with a toolbox placed on it. The force of gravity acting on the toolbox is vertically downward and has a magnitude of 70 lb. The force required to hold the toolbox in place is acting parallel to the incline and has a magnitude of 36 lb. We need to find the angle that the incline makes with the horizontal.
Now, let's resolve the weight of the toolbox into its components. The weight can be split into two directions: perpendicular (or normal) to the incline and parallel to the incline. The component of the weight perpendicular to the incline is given by W⊥ = W * cosθ, where θ is the angle we want to find. In this case, W refers to the weight of the toolbox, which is 70 lb. So, W⊥ = 70 * cosθ.
The force required to hold the toolbox in place is acting perpendicular to the incline, which we can call F⊥. This force balances out the perpendicular component of the weight, so F⊥ = W⊥. Thus, F⊥ = 70 * cosθ.
Since we know that the force required to hold the toolbox in place is 36 lb, we can set up an equation: 36 = 70 * cosθ.
Now, we can solve for θ. Divide both sides of the equation by 70: cosθ = 36/70. Then, take the inverse cosine (cos⁻¹) of both sides: θ = cos⁻¹(36/70).
Using a calculator, you can input 36/70, take the inverse cosine, and get the answer for θ.
Therefore, the angle that the incline makes with the horizontal is θ = cos⁻¹(36/70).