# Problem Breakdown We are given: Weight lifter mass: \( m_{\text{person}} = 74.5\,\text{kg} \) - Weight in hand: \( m_{\text{weight}} = 7.50\,\text{kg} \) - Forearm+hand mass: \( m_{\text{fh}} = 5.70\% \) of total body mass - Distance from elbow to: - biceps attachment: \( r_{\text{biceps}} = .055\,\text{m} \) - forearm-hand center of mass: \( r_{\text{fh}} = .170\,\text{m} \) - weight: \( r_{\text{weight}} = .380\,\text{m} \) - Arm is 40° below horizontal (angle with respect to horizontal: \(\theta = 40^\circ\)) Let’s use \( g = 9.8\,\text{m/s}^2 \). --- ## 1. **Force the Biceps Muscle Exerts on the Forearm** ### Step 1: Analyze Torques About the Elbow Joint **Set up the torque equilibrium equation (taking counterclockwise as positive):** \[ \sum \tau = \] The forces causing torque about the elbow: - **Biceps force (\( F_B \))** (up, at \( r_{\text{biceps}} \), vertical) - **Forearm-hand weight (\( W_{fh} \))** (down, at \( r_{\text{fh}} \), acts vertically down) - **External weight (\( W \))** (down, at \( r_{\text{weight}} \), acts vertically down) All forces act at \( \theta = 40^\circ \) with respect to the lever arm. \[ F_B \cdot r_{\text{biceps}} \cdot \sin(40^\circ) = W_{fh} \cdot r_{\text{fh}} \cdot \sin(40^\circ) + W \cdot r_{\text{weight}} \cdot \sin(40^\circ) \] Since all torques have the same \(\sin(40^\circ)\), it cancels: \[ F_B \cdot r_{\text{biceps}} = W_{fh} \cdot r_{\text{fh}} + W \cdot r_{\text{weight}} \] ### Step 2: Plug in the Numbers - \( m_{\text{fh}} = .057 \times 74.5 = 4.2465\,\text{kg} \) - \( W_{fh} = 4.2465 \times 9.8 = 41.62\,\text{N} \) - \( W = 7.50 \times 9.8 = 73.5\,\text{N} \) Plug into the equation: \[ F_B \cdot .055 = 41.62 \cdot .170 + 73.5 \cdot .380 \] \[ F_B \cdot .055 = 7.075 + 27.93 = 35.005 \] \[ F_B = \frac{35.005}{.055} = 636.5\,\text{N} \] **Explanation:** We sum the torques due to each force about the elbow. The biceps must provide enough upward force to balance the downward torques from the weight and the arm's mass. **Final Answer for Step 1:** \[ \boxed{F_B = 637\,\text{N}} \] (Rounded to 3 significant figures) --- ## 2. **Magnitude and Direction of the Elbow Joint Force** ### Step 1: Find Vertical and Horizontal Force Components #### **Vertical Forces:** \[ \sum F_y = \] \[ F_{B} + F_{E,y} = W_{fh} + W \] \[ F_{E,y} = W_{fh} + W - F_B \] \[ = 41.62 + 73.5 - 636.5 = -521.38\,\text{N} \] #### **Horizontal Forces:** Biceps pulls vertically; the weights act vertically, so the horizontal component is zero unless the biceps pulls at an angle. But the problem says the biceps acts vertically, so: \[ \sum F_x = \implies F_{E,x} = \] **Explanation:** The elbow joint must provide a strong upward force to balance the weights, but since the biceps pulls upward and the weights pull downward, the elbow force comes out negative (i.e., directed downward). ### Step 2: Magnitude and Direction - Magnitude: \( |F_{E}| = |F_{E,y}| = 521\,\text{N} \) - Direction: **Downward** (since the value is negative) **Final Answer for Step 2:** - **Magnitude:** \(\boxed{521\,\text{N}}\) - **Direction:** \(\boxed{\text{Downward}}\) --- ## 3. **Effect of Raising Arm Toward Horizontal** As the arm becomes more horizontal, the perpendicular moment arm for the weights increases (since \(\sin\theta\) increases toward 1 at \(90^\circ\)), so the torque due to the weights increases. The biceps must exert a **larger force** to balance this. **Final Answer:** - **The force in the biceps muscle will \(\boxed{\text{increase}}\) as the arm approaches the horizontal.**