The following is a simple method of measuring coefficients of friction: Suppose a block is placed on a rough surface inclined relative to the horizontal, as shown in Figure 5.19. The incline angle is increased until the block starts to move. Show that by measuring the critical angle θc at which

Question

The following is a simple method of measuring coefficients of friction: Suppose a block is placed on a rough surface inclined relative to the horizontal, as shown in Figure 5.19. The incline angle is increased until the block starts to move. Show that by measuring the critical angle [latex] heta_{c} [/latex] at which this slipping just occurs, we can obtain [latex] \mu_{s} [/latex] .

Accepted Answer

Conceptualizing from the free body diagram in Figure 5.19, we see that we can categorize this as a Newton’s second law problem. To analyze the problem, note that the only forces acting on the block are the gravitational force mg, the normal force n, and the force of static friction [latex] f _{s } [/latex] . These forces balance when the block is not moving. When we choose x to be parallel to the plane and y perpendicular to it, Newton’s second law applied to the block for this balanced situation gives

[latex] ext { (1) } \quad \sum F_{x}=m g \sin heta-f_{s}=m a_{x}=0 [/latex].

[latex] ext { (2) } \quad \sum F_{y}=n-m g \cos heta=m a_{y}=0 [/latex].

[latex] ext { We can eliminate } m g ext { by substituting } m g=n / \cos heta ext { from (2) } [/latex] into (1) to find

[latex] ext { (3) } \quad f_{s}=m g \sin heta=\left(\frac{n}{\cos heta} ight) \sin heta=n an heta [/latex].

When the incline angle is increased until the block is on the verge of slipping, the force of static friction has reached its maximum value [latex] \mu_{s} n . ext { The angle } heta [/latex] in this situation is the critical angle [latex] heta_{c}, ext { and (3) becomes } [/latex]

[latex] \mu_{s} n=n an heta_{c} [/latex]

[latex] \mu_{s}= an heta_{c} [/latex].

For example, if the block just slips at [latex] heta_{c}=20.0^{\circ} [/latex] , then we
find that [latex] \mu_{s}= an 20.0^{\circ}=0.364 [/latex].

To finalize the problem, note that once the block starts to move at [latex] heta \geq heta_{c}, [/latex] it accelerates down the incline and the force of friction is [latex] f_{k}=\mu_{k} n . ext { However, if } heta [/latex] is reduced to a value less than [latex] heta_{c} ext {, it may be possible to find an angle } heta_{c}^{\prime} [/latex] such that the block moves down the incline with constant speed [latex] \left(a_{x}=0 ight) [/latex] In this case, using (1) and (2) with [latex] f_{s} ext { replaced by } f_{k} ext { gives } [/latex]

[latex] \mu_{k}= an heta_{c}^{\prime} [/latex].

[latex] ext { where } heta_{c}^{\prime}< heta_{c} [/latex].

Question 5.12: The following is a simple method of measuring coefficients o...

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