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SOVA2 [1]
3 years ago
13

A refrigerator is being pulled up a ramp with a horizontal force P, which acts at the top corner. The refrigerator has a mass of

75 kg, acting through point G. The ramp is inclined at 20º, and the coefficient of static friction is 0.3 between the refrigerator and the ramp.
(a) Find the force P required to move the refrigerator.
(b) Does the refrigerator tip or slide?
Physics
1 answer:
vaieri [72.5K]3 years ago
4 0

Answer:

(a) P = 459.055 N.

(b) the refrigerator tips.

Explanation:

Given, the angle of ramp is 20°.

When the weight of refrigerator is resolved in directions parallel and perpendicular to ramp, 75×g×sin(20°) and 75×g×cos(20°).

⇒ normal contact force is 75×g×cos(20°).

⇒ frictional force is 0.3×75×g×cos(20°) = 207.414 N

so, total opposite force is 207.414 + 75×g×sin(20°) = 459.055 N.

so, the force needed is P = 459.055 N

And as the moment due to both opposite force and P force are in same direction the refrigerator tips rather than just sliding.

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<h2>a) Q = 0.759µC</h2><h2>b) E = 39.5µJ</h2>

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b) Energy stored in a capacitor is expressed as E = 1/2CV²

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Explanation:

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F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

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