Question

a block of mass m = 2.2 kg sits at rest in the middle of an incline that makes an angle θ with the horizontal. the coefficient of static friction between the block and the incline is μₛ = 0.400 and the coefficient of kinetic friction is μₖ = 0.300. the angle θ can be varied between θ = 0° and θ = 45°. initially the angle is θ = 0° (so that the incline is parallel to the ground) and then it is slowly increased until it reaches θ = 45°. at what angle θ does the block begin to slide down the incline?
○ 21.8°
○ 16.7°
○ 23.6°
○ 17.46°
○ the block never slides down the incline.

Explanation:

Step1: Set up equilibrium condition

The block starts to slide when the component of gravity along the incline equals the maximum static friction force. The formula is:
$$mg\sin\theta = \mu_s mg\cos\theta$$

Step2: Simplify the equation

Cancel $mg$ from both sides, then rearrange to solve for $\theta$:
$$\tan\theta = \mu_s$$

Step3: Substitute value and calculate

Substitute $\mu_s = 0.400$, then find $\theta$ using the arctangent function:
$$\theta = \arctan(0.400)$$

Answer:

21.8°