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MariettaO [177]
3 years ago
5

Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If

the coefficient of static friction between the brick and the floor is 0.4, what is the minimum value of F required to start the brick to move?
Physics
1 answer:
jeka57 [31]3 years ago
7 0

Let F_{1}=F_{2}=F.

Normal force equals (using Newton's third law) N=mg+F\sin30^{o}-F\sin37^{o}.

F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o}), but F_{f}\leq F(\cos30^o+\cos37^o) for all F_{f} (in order to start moving the break). Therefore F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o}), solving for F: F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}

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Explanation:

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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
weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

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:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A   (N/m²)

P1=P2

\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

A₁ = 2.0 cm²

A₂ = 40 cm²

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

We replace data in the equation (1)

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

F₂ = 10000 N

W₂= F₂= 10000 N

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Answer:

B . energy cannot be created or destroyed

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Gekata [30.6K]

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To learn more about Internal energy -

<u>brainly.com/question/11623849</u>

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