Explanation Given information: The weight of the block is W = 200 N . The magnitude of the force P is 100 N. Calculation: Draw the free-body diagram of the block as in Figure (1). Use Figure (1), Resolve the vertical component of forces. ∑ F y = 0 − 200 + R 2 sin ϕ s = 0 R 2 sin ϕ s = 200 Draw the free-body diagram of the wedge and force triangle as in Figure (2). Find the value of angle ϕ s using the sine rule. 100 = 200 tan ϕ s + 200 tan ( ϕ s + 15 ° ) (1) Solve the Equation (1) by trial and error procedure. First trial: Try ϕ s = 6 ° : Find the value of angle ϕ s : Substitute 6 ° for ϕ s . 100 = 200 tan ϕ s + 200 tan ( ϕ s + 15 ° ) 100 = 200 tan 6 ° + 200 tan ( 6 ° + 15 ° ) 100 ≠ 97.80 Hence, the assumption is incorrect. Second trial: Try ϕ s = 6.3 ° : Find the value of angle ϕ s : Substitute 6

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Chapter 8.2, Problem 8.66P

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The coefficient of static friction (μs) for which the motion is impending.

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Chapter 8.2, Problem 8.65P

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Chapter 8 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

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