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lakkis [162]
3 years ago
11

Bob is pulling a 30kg filing cabinet with a force of 200N , but the filing cabinet refuses to move. The coefficient of static fr

iction between the filing cabinet and the floor is 0.80.
What is the magnitude of the friction force on the filing cabinet?
Physics
1 answer:
Lostsunrise [7]3 years ago
5 0

Answer:

Magnitude force of friction is 235.2 N

Explanation:

The force of friction is the product of weight and the coefficient of static friction

The force have to be bigger than 200N because it don't move

∑F= F_{n} - F_{fr}

F_{fr} = u*m*g\\F_{fr} = 0.80*30kg*9.8 \frac{m}{s^{2} } \\F_{fr} = 235.2 \frac{kg*m}{s^{2} } \\F_{fr} = 235.2 } N

The force on the filling cabinet

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A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area
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Answer:

W₂= 10000 N

Explanation:

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If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

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\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

F_{2} = \frac{F_{1}*A_{2}  }{A_{1}}  Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

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Calculation of the weight  (W₂) can the large piston support

We replace data in the equation (1)

F_{2} = \frac{(500)*(40) }{2}

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