Two crates connected by a rope lie on a horizontal surface (Fig. E5.375.375.37). Crate A has mass mAm_AmA, and crate B has mass mBm_BmB. The coefficient of kinetic friction between each crate and the surface is μkμ_kμk. The crates are pulled to the right at constant velocity by a horizontal force FFF. Draw one or more free-body diagrams to calculate the following in terms of mAm_AmA, mBm_BmB, and μkμ_kμk: the tension in the rope connecting the blocks.

Ch 05: Applying Newton's Laws

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 05: Applying Newton's Laws

문제 37b

5장, 문제 37b

Two crates connected by a rope lie on a horizontal surface (Fig. E $5.37$ ). Crate A has mass $m_A$ , and crate B has mass $m_B$ . The coefficient of kinetic friction between each crate and the surface is $μ_k$ . The crates are pulled to the right at constant velocity by a horizontal force $F$ . Draw one or more free-body diagrams to calculate the following in terms of $m_A$ , $m_B$ , and $μ_k$ : the tension in the rope connecting the blocks.

검증된 단계별 안내

Step 1: Begin by analyzing the forces acting on each crate. For crate A (mass mA), the forces include the applied force F, the tension T in the rope connecting it to crate B, the kinetic friction force opposing motion (μk * mA * g), and the normal force balancing the weight of the crate (mA * g). For crate B (mass mB), the forces include the tension T pulling it to the right, the kinetic friction force opposing motion (μk * mB * g), and the normal force balancing the weight of the crate (mB * g).

Step 2: Draw free-body diagrams for both crates. For crate A, represent the applied force F pointing to the right, the tension T pointing to the left, and the friction force μk * mA * g pointing to the left. For crate B, represent the tension T pointing to the right and the friction force μk * mB * g pointing to the left.

Step 3: Use Newton's second law (ΣF = ma) for each crate. Since the crates are moving at constant velocity, the acceleration is zero, and the net force on each crate is zero. For crate A, the equation becomes F - T - μk * mA * g = 0. For crate B, the equation becomes T - μk * mB * g = 0.

Step 4: Solve the equation for crate B to find the tension T. Rearrange the equation T - μk * mB * g = 0 to get T = μk * mB * g.

Step 5: Substitute the value of T from crate B into the equation for crate A to verify consistency. The tension in the rope connecting the blocks is determined by the frictional force acting on crate B, which depends on its mass mB, the coefficient of kinetic friction μk, and the acceleration due to gravity g.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Free-Body Diagram

A free-body diagram is a graphical representation used to visualize the forces acting on an object. In this scenario, each crate (A and B) will have its own free-body diagram showing the applied force, tension in the rope, and frictional forces. This helps in analyzing the forces to determine the net force and subsequently the tension in the connecting rope.

Kinetic Friction

Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. It is quantified by the coefficient of kinetic friction (μk) multiplied by the normal force. In this problem, the frictional force acting on each crate must be considered to calculate the net force and the tension in the rope, as it affects the overall motion of the system.

Newton's Second Law

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). In this case, since the crates are moving at constant velocity, the net force is zero, which means the tension in the rope must balance the frictional forces acting on both crates. This principle is essential for calculating the tension in the rope.