Chapter 12: Q55P (page 350)

In Fig. 12-64, block A $(m a s s 10 k g)$ is in equilibrium, but itwould slip if block B $(m a s s 5 .0 k g)$ were any heavier. For angle $θ = 30 °$ what is the coefficient of static friction between block Aand the surfacebelow it?

Short Answer

Expert verified

The coefficient of static friction between block A and the surface is $0.288$

Step by step solution

Listing the given quantities

$m_{A} = 10 k g$

$m_{B} = 5 k g$

$θ = 30 °$

Understanding the concept of coefficient of static friction

By drawing the free body diagram of the situation, we can find the tension in each wire employing which they are connected. Using the conditions for equilibrium, we can find the coefficient of static friction.

Formula:

$\sum_{}^{} F_{y} = 0$

$\sum_{}^{} F_{x} = 0$

$F_{s t a t i c f r i c t i o n} = μ_{s} \times N o r m a l F o r c e$

Free body diagram block A

From this, we can say that,

$T_{1} = F_{s}$

$T_{1} = μ_{s} \times F_{N}$

$T_{1} = μ_{s} \times (m_{A} \times g)$

$T_{1} = μ_{s} \times (10 kg \times 9.8 m / s^{2})$

$T_{1} = 98 N \times μ_{s}$

Free body diagram block B

From this, we can say that,

$T_{2} = m_{B} \times g$

$\begin{array}{l} T_{2} = 5 k g \times 9.8 m / s^{2} \\ T_{2} = 49 N \end{array}$

Free body diagram for equilibrium position

Calculations of static friction

From this, we can say that,

$\begin{array}{l} \sum_{}^{} F_{y} = 0 \\ (T_{3} ×cosθ) {-T}_{2} = 0 \\ (T_{3} ×cosθ) = T_{2} \\ (T_{3} ×cosθ) = 49 N \end{array}$

$\begin{array}{l} \sum_{}^{} F_{x} = 0 \\ (T_{3} \times sinθ) {-T}_{1} = 0 \\ (T_{3} \times sinθ) = T_{1} \\ (T_{3} \times sinθ) = 98 N \times μ_{s} \end{array}$

$\begin{array}{l} \frac{(T_{3} \times sinθ)}{(T_{3} \times cosθ)} = \frac{98 N \times μ_{s}}{49 N} \\ μ_{s} = \tan θ \times 0.5 \\ μ_{s} = 0.5 \times \tan (30 °) \\ μ_{s} = 0.5 \times 0.577 \\ μ_{s} = 0.288 \end{array}$

The coefficient of static friction between block A and the surface is $0.288$

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In Fig. 12-64, block A $(m a s s 10 k g)$ is in equilibrium, but itwould slip if block B $(m a s s 5 .0 k g)$ were any heavier. For angle $θ = 30 °$ what is the coefficient of static friction between block Aand the surfacebelow it?

Short Answer

Step by step solution

Listing the given quantities

Understanding the concept of coefficient of static friction

Free body diagram block A

Free body diagram block B

Free body diagram for equilibrium position

Calculations of static friction

One App. One Place for Learning.

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In Fig. 12-64, block A (mass 10 kg)is in equilibrium, but itwould slip if block B (mass 5.0 kg)were any heavier. For angle θ=30°what is the coefficient of static friction between block Aand the surfacebelow it?

Short Answer

Step by step solution

Listing the given quantities

Understanding the concept of coefficient of static friction

Free body diagram block A

Free body diagram block B

Free body diagram for equilibrium position

Calculations of static friction

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Torque and Rotational Motion

Thermodynamics

Translational Dynamics

Physics of Motion

Fundamentals of Physics

Study anywhere. Anytime. Across all devices.

In Fig. 12-64, block A $(m a s s 10 k g)$ is in equilibrium, but itwould slip if block B $(m a s s 5 .0 k g)$ were any heavier. For angle $θ = 30 °$ what is the coefficient of static friction between block Aand the surfacebelow it?