Question

A 5 kg block rests on an inclined plane with a coefficient of static friction equal to 0.30. What is the minimum angle at which the block will begin to slide

Answer 1

Answer:

$\theta = 16.70 ^{\circ}$

Explanation:

The coefficient of static friction is equal to the tangent of the minimum angle at which an object will begin to start sliding down a ramp.  

• $\displaystyle u_s=\frac{F_f}{F_N} = \frac{F_g\ \text{sin}\theta}{F_g\ \text{cos} \theta} = \text{tan} \theta$

Since we are given the coefficient of static friction we can solve for the minimum angle that the block will begin to slide.

Let's solve for the force of gravity that is acting on the block. The force of gravity is also known as the weight force, which can be calculated by using w = mg.

• $w=mg$

We are given the mass of the block (kg) and we know that g = 9.8 m/s².

• $w=(5)(9.8) = 49 \ \text{N}$

Now we can use this force in the equation:

• $\displaystyle u_s = \frac{F_g \ \text{sin} \theta }{F_g \ \text{cos} \theta}$

Plug $\displaystyle u_s = 0.30$ and 49 N into the equation.

• $\displaystyle 0.30 = \frac{(49) \ \text{sin} \theta }{(49) \ \text{cos} \theta}$  
• $0.30=\text{tan} \theta$

Notice that the gravitational force cancels out in the end, so we can actually start with $0.30=\text{tan} \theta$.

Evaluate this equation by taking the inverse tangent of both sides of the equation.

• $\text{tan}^-^1 (0.30) = \text{tan}^-^1 (\text{tan}\theta)$
• $\text{tan}^-^1 (0.30) =\theta$
• $\theta = 16.69924423$

The minimum angle at which the block will begin to slide is about 16.70 degrees.