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Required information
Block $A$ supports a pipe column and rests as shown on wedge $B$. The coefficient of static friction at all surfaces of contact is 0.25 and that $ heta=45^{\circ}$. Assume the wedges are massless.

Determine the smallest force P required to raise block $A$.
Given: $W=7 \mathrm{kN}$.
The smallest force $\mathbf{P}$ required to raise block $A$ is $\square$.

Question

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Required information
Block $A$ supports a pipe column and rests as shown on wedge $B$. The coefficient of static friction at all surfaces of contact is 0.25 and that $	heta=45^{\circ}$. Assume the wedges are massless.

Determine the smallest force P required to raise block $A$.
Given: $W=7 \mathrm{kN}$.
The smallest force $\mathbf{P}$ required to raise block $A$ is $\square$.

Accepted Answer

∻Solution∻ ‖ a ⋮ Start by analyzing the forces acting on block A. The weight of block A is given as $$W = 7 \, ext{kN}$$ acting downward. The static friction force, which opposes the motion, can be expressed as $$f_{ ext{static}} = \mu_s N$$, where $$\mu_s = 0.25$$ is the coefficient of static friction and $$N$$ is the normal force exerted by wedge B on block A. Since the wedge is massless, the normal force is equal to the vertical component of the force exerted by the wedge. ‖ ‖ b ⋮ The normal force $$N$$ can be resolved into components. The vertical component of the force due to the weight of block A is $$W$$, and the horizontal component is influenced by the applied force $$P$$. The angle $$ heta$$ is given as $$45^{\circ}$$, so we can express the normal force as $$N = P \cos(45^{\circ})$$. The vertical equilibrium condition gives us: $$N \sin(45^{\circ}) = W$$. Substituting for $$N$$, we have: $$P \cos(45^{\circ}) \sin(45^{\circ}) = W$$. ‖ ‖ c ⋮ Since $$\cos(45^{\circ}) = \sin(45^{\circ}) = \frac{\sqrt{2}}{2}$$, we can simplify the equation to: $$P \left(\frac{\sqrt{2}}{2} ight)^2 = W$$, which leads to: $$P \cdot \frac{1}{2} = 7 \, ext{kN}$$. Therefore, we find that $$P = 14 \, ext{kN}$$. ‖ ∻Answer∻ ⚹ 14 kN ⚹ ∻Key Concept∻ ⚹ Static Equilibrium: In static equilibrium, the sum of forces acting on an object must be zero. This includes resolving forces into their components and applying the conditions for equilibrium. The frictional force is also a key factor in determining the forces required to maintain equilibrium. ⚹ ∻Explanation∻ ⚹ The smallest force P required to raise block A is determined by balancing the forces acting on it, considering both the weight and the frictional forces involved. By applying the equilibrium conditions, we find that P must be 14 kN to overcome the weight of block A. ⚹

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