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Answer: 37 kg (approx.) The acceleration of gravity on the moon is 1/6-th the value of earth's; this explains why **the force** of gravity is 1/6-th that of earth's. **The force** of gravity (**F** grav) on the moon is calculated using the equation **F** grav = m•g where g =1.6 m/s/s (on the moon). If the magnitude of the **force**, **F**, that you exert on the box is small, the frictional **force**, **f**, acting in the direction opposing **F**, will **keep** **the** box from moving, and the box remains in a state of static **equilibrium**. If the magnitude of the **force**, **F**, **is** gradually increased, the box will begin to move across the ﬂoor. Once the box begins. **The Force F Needed to Keep the Block at Equilibrium (pulley and** String are massless) Dear Simran Question is incomplete as there is no image attached to solve. ×. [Click Here for Sample Questions] 1. When two forces are equal and oppositely directed, they are in **Equilibrium**. 2. The vectors are not parallel to each other and do not meet at a location for non-concurrent coplanar forces. 3. Three parallel coplanar forces are in balance. 4. It is impossible to achieve **Equilibrium** when only a single **f**.

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**Given** that the coefficient of kinetic friction between the crate and the ground floor is 0.100 and between the crate and the third floor surface is 0.300, find the work **needed** by the workers for each path shown from A to E. Assume that **the force** the workers need to do is just enough to slide the crate at constant velocity (zero acceleration).

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**force**, an upward **force** applied on each **block** by the fluid. **Figure** 9.5: In this case, the **blocks** are not at **equilibrium**. **The** **block** on the left has been pushed down into the water and released. Because it displaces more water than it does at **equilibrium**, **the** buoyant **force** applied to it by the water is larger than the **force** of gravity. .

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From this FBD, we have the following **equilibrium** equations: X **F** x= P **f**= 0 ! **f**= P X **F** y= N W= 0 ! N= W From the rst equation, we see that, as long as **the block** remains in **equilibrium**, the friction **force** is always equal to the applied **force** P. However, from the above discussion, we know that the friction **force** has a maximum value of **f** max = sN.

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. **Force** Analysis of Rolling Body • Resultant of distributed normal **force** • To **keep** **the** cylinder in **equilibrium**, all the **forces** must be concurrent. • Resultant **force** will pass through the center and making an angle of ɵ with vertical • Taking a moment about A, we get Assuming small ɵ, cos (ɵ) ≈ 1 𝑊𝑎=𝑃𝑟cos𝜃 𝑊𝑎 𝑟.

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Solution: Chapter 11 Rotational Dynamics and Static **Equilibrium** Q.61P. Two gerbils run in place with a linear speed of 0.55 m/s on an exercise wheel that is shaped like a hoop. Find the angular momentum of the system if the gerbil has a mass of 0.22 kg and the exercise wheel has a radius of 9.5 cm and a mass of 5.0 g. **The** x and y components of **F** 2 and **F** 3 cancel out, leaving only **F** 1 as the net **force**. So the work done is **F** 1d = 12 J 2. The **figure** gives the acceleration of a 4 kg object as an applied **force** moves it from rest along an x-axis from x=0 to x=9 m. The scale of the **figure's** vertical axis is set by as=10.0 m/s^2. What is the object's speed when it. The two **blocks** shown above are sliding across a frictionless surface by a **force F** from the left. The two **blocks** are not attached but the coefficient of static friction between the two is μ s = 0.37. The mass of the smaller **block** is m 1 = 19.0 kg and the mass of the larger **block** is m 2 = 85.0 kg. What minimum **force F** is **needed to keep** M 1 from. (This contains all **needed** information according to my professor) A **block** of mass m = 2.61 kg is held in **equilibrium** on a frictionless incline of angle θ = 56.7 by the horizontal **force** **F**, as shown in the **figure**. (a) Determine the value of **F** (N). (b) Determine the normal **force** (N) exerted by the incline on the **block**.

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- 5.Draw the shear
**force**and bending moment diagrams for the simple supported beam as shown in Fig. Ans: For**equilibrium**conditions ∑**f**=0, R B+R**F**= 70 Taking moment about point**'F'**R B x 8 = 40 x 10 + 20 x I x 7.5 + 10 x 7 + 20 x 2 x 6 + 20 x 5 + 20 x 2 + 10 x 2 x 1 R B = 127.5 kN R**F**= 42.5 kN Shear**force**diagram: - The force f needed to keep the block at equilibrium in given figure is The
**figure**shows a pulley**block**system placed on a smooth table. All pulleys are light and smooth and the strings are light. The mass of each block is 5 k g and the angle θ = 6 0 ∘. (Take g = 1 0 m / s e c 2). The acceleration of hanging block just after the. Hint 1. - The formula for Hooke’s law specifically relates the change in extension of the spring, x , to the restoring
**force**,**F**, generated in it: The extra term, k , is the**spring constant**. The value of this constant depends on the qualities of the specific spring, and this can be directly derived from the properties of the spring if**needed**. **Block**will be in**equilibrium**and friction will act in downward direction having magnitude 23.46lb. a) minimum**force**required to start**block**up is 74.8lb. b) minimum**force**required to**keep**it moving is 59.357lb. c) minimum**force**required to prevent it from moving down is 6.75 lb- The forces on this
**block**are shown in**figure**(6-W16) with M = 2 kg. It is assumed that m has its minimum value so that the 2 kg**block**has a tendency to slip down. As**the block**is in**equilibrium**, the resultant**force**should be zero. Taking components ⊥ to the incline N = Mg cos 45 ° = Mg/ √ 2. Taking components | | to the incline T +**f**= Mg ...