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Friction Additional Examples

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Example 1: A 100-lb force acts as shown on a 300-lb block placed on an inclined plane. The coefficients of friction between the block and the plane are ms =0.25 and mk = Determine whether the block is in equilibrium, and find the value of the friction force.

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**Force required for equilibrium: Assuming that F is directed down**

SFx = 100lb – 3/5 (300lb) – F = 0 F = 80 lb SFy = N – 4/5(300lb) = 0, N = +240 lb The F required to maintain equilibrium is directed up and to the right; the tendency of the block is move down the plane. 2. Maximum friction force. Fm = ms N, Fm = 0.25(240 lb) = 60 lb Since the value of F required to maintain equilibrium (80lb) is larger than the maximum possible (60lb), the block will slide down the plane

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**3. Actual value of friction force:**

Factual = Fk ( the body is moving) Factual = Fk = mk N = 0.2(240lb) = 48 lb The sense of this force is opposite to the sense of motion. The forces acting on the block are not balanced, the resultant is: 3/5 (300lb) – 100lb – 48lb = 32lb

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Example2: Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when q = 30o and P = 50lb.

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**y x q q N F Assume equilibrium: SFy =N – 250cos30o-50sin30o = 0**

N = lb SFx =F– 250sin30o +50cos30o = 0, F = lb x q q N F 2. Maximum friction force: Fm = ms N = 0.3 (241.5 lb) = 72.5 lb Since F > Fm, then the block moves down Friction force: F = mk N = 0.2(241.5lb) = 48.3 lb

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**Example 3: The block in the figure has a mass of 100 kg**

Example 3: The block in the figure has a mass of 100 kg. The coefficient of friction between the block and the inclined surface is 0.2. (a) Determine if the system is in equilibrium when P= 600 N P 20o 30o

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**W= 981 N y P x 30o 20o F N 30o Determination of F and N:**

SFy= Psin 20o + N – Wcos 30o = 0, N = N SFx= Pcos 20o – F – Wsin 30o = 0, F = N 2. Determination of Fmaximum Fm =msN= (0.20)(644.36)=128.87N Since Fm > 73.32, then the block is in equilibrium. P x 30o 20o F N 30o

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**b) Determine the minimum force P to prevent motion**

y Pmin The minimum P will be required when motion of the block down the incline is impending. F must resist this motion as shown. Equilibrium exists when: SFx = Pmin cos 20o + F – 981sin 30o = 0 SFx = Pmin cos 20o + 0.2N – 981 sin 30o = 0 SFy = Pmin sin 20o + N – 981cos 30o = 0 Then N = 724 N, P min = 368 N x 30o 20o F N 30o

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**c) Determine the maximum force P for which the system is in equilibrium**

The maximum force P will be required when motion of the block up the incline is impending. For this condition, F will tend to resist this motion as shown. Then: SFx = Pmax cos 20o – 0.2 N – 981 sin 30o = 0 SFy = Pmax sin 20o + N – 981 cos 30o = 0 Solving simultaneously: N = 626 N Pmax = 655N 981 N y Pmax x 30o 20o F N 30o

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Center of Gravity Additional Examples

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**Centroids – Simple Example for a Composite Body**

Find the centroid of the given body

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**Centroids – Simple Example for a Composite Body**

To find the centroid, Determine the area of the components

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**Centroids – Simple Example for a Composite Body**

The total area is

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**Centroids – Simple Example for a Composite Body**

To centroid of each component Compute the x centroid

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**Centroids – Simple Example for a Composite Body**

To centroid of each component Compute the y centroid

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**Centroids – Simple Example for a Composite Body**

The problem can be done using a table to represent the composite body.

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**Centroids – Simple Example for a Composite Body**

An alternative method of computing the centroid is to subtract areas from a total area. Assume that area is Assume that area is a large square and subtract the small triangular area.

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**Centroids – Simple Example for a Composite Body**

The problem can be done using a table to represent the composite body.

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**Centroids –Example for a Composite Body**

Find the centroid of the given body

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**Centroids –Example for a Composite Body**

Determine the area of the components

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**Centroids –Example for a Composite Body**

The total area is

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**Centroids – Example for a Composite Body**

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**Centroids – Example for a Composite Body**

An Alternative Method would be to subtract to areas

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**Centroids – Class Problem**

Find the centroid of the body

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**Centroids – Class Problem**

Find the centroid of the body

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Friction Ffriction = μFNormal.

Friction Ffriction = μFNormal.

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