Question

1. a block is on a horizontal frictionless surface. an applied force of magnitude $f_a$ is exerted on the block at an angle $\theta$ to the horizontal as shown in the figure below, causing the block to accelerate to the right. the surface exerts a normal force of magnitude $f_n$ on the block. while $\theta$ is kept constant, the magnitude of $f_a$ is doubled. assume the block remains in contact with the surface. how does the new normal force exerted on the block compare to $f_n$? (greater, less, the same, or more information is needed) justify your answer.

1. if the tension in $t_1$ and $t_2$ is 200 n, find the tension in $t_3$.

1. determine the tension in both of the strings attached to the light

1. determine the tension in each string. the angle between $t_2$ and the ceiling is $45^\circ$.

Explanation:

Problem 8

Step1: Analyze Vertical Forces

The block has weight \( mg \) downward, normal force \( F_N \) upward, and vertical component of \( F_A \) upward (\( F_A \sin\theta \)). So vertical equilibrium: \( F_N + F_A \sin\theta = mg \), so \( F_N = mg - F_A \sin\theta \).

Step2: Double \( F_A \)

New \( F_A' = 2F_A \). New normal force \( F_N' = mg - F_A' \sin\theta = mg - 2F_A \sin\theta \).

Step3: Compare \( F_N' \) and \( F_N \)

Since \( F_N = mg - F_A \sin\theta \) and \( F_N' = mg - 2F_A \sin\theta \), \( F_N' = F_N - F_A \sin\theta \). Thus, \( F_N' < F_N \) (because \( F_A \sin\theta > 0 \)).

Step1: Analyze Horizontal Forces

\( T_1 \) and \( T_2 \) have horizontal components \( T_1 \cos40^\circ \) (left) and \( T_2 \cos40^\circ \) (right). Since \( T_1 = T_2 = 200\,\text{N} \), horizontal forces balance.

Step2: Analyze Vertical Forces

Vertical components of \( T_1 \) and \( T_2 \) are \( T_1 \sin40^\circ \) and \( T_2 \sin40^\circ \), upward. \( T_3 \) is downward (supports mass \( m \)). So \( T_1 \sin40^\circ + T_2 \sin40^\circ = T_3 \).

Step3: Calculate \( T_3 \)

Substitute \( T_1 = T_2 = 200\,\text{N} \): \( T_3 = 200\sin40^\circ + 200\sin40^\circ = 400\sin40^\circ \approx 400 \times 0.6428 \approx 257.1\,\text{N} \).

Step1: Identify Forces

Let tension in the slanted string be \( T_1 \), horizontal string be \( T_2 \), weight \( 100\,\text{N} \) downward. Vertical equilibrium: \( T_1 \sin37^\circ = 100\,\text{N} \). Horizontal equilibrium: \( T_1 \cos37^\circ = T_2 \).

Step2: Find \( T_1 \)

From \( T_1 \sin37^\circ = 100 \), \( T_1 = \frac{100}{\sin37^\circ} \approx \frac{100}{0.6} \approx 166.7\,\text{N} \).

Step3: Find \( T_2 \)

\( T_2 = T_1 \cos37^\circ \approx 166.7 \times 0.8 \approx 133.3\,\text{N} \).

Answer:

less

Problem 9