please show or justify all answers. work must be shown for full credit.
8. while doing bicep curls, tamara applies 155 newtons of force to lift the dumbbell. her forearm is 0.366 meters long and she begins the bicep curl with her elbow bent at a 15° angle below the horizontal, in the direction of the positive x - axis. determine the magnitude of the torque about her elbow.
(show work)

9. jenny is sitting on a sled on the side of a hill inclined at 15°. what force is required to keep the sled from sliding down the hill if the combined weight of jenny and the sled is 90 pounds?
(show work)

10. anne is pushing a wheelbarrow filled with mulch to place in her garden. she is pushing the wheelbarrow with a force of 70 n at an angle of 50° with the horizontal. how much work in joules is anne doing when she pushes the wheelbarrow 25 meters?
(show work)

11. a baseball player running forward at 3 meters per second throws a ball with a velocity of 20 meters per second at an angle of 20° with the horizontal. what is the resultant speed and direction of the throw?
(show work)

Question

please show or justify all answers. work must be shown for full credit.
8. while doing bicep curls, tamara applies 155 newtons of force to lift the dumbbell. her forearm is 0.366 meters long and she begins the bicep curl with her elbow bent at a 15° angle below the horizontal, in the direction of the positive x - axis. determine the magnitude of the torque about her elbow.
(show work)

9. jenny is sitting on a sled on the side of a hill inclined at 15°. what force is required to keep the sled from sliding down the hill if the combined weight of jenny and the sled is 90 pounds?
(show work)

10. anne is pushing a wheelbarrow filled with mulch to place in her garden. she is pushing the wheelbarrow with a force of 70 n at an angle of 50° with the horizontal. how much work in joules is anne doing when she pushes the wheelbarrow 25 meters?
(show work)

11. a baseball player running forward at 3 meters per second throws a ball with a velocity of 20 meters per second at an angle of 20° with the horizontal. what is the resultant speed and direction of the throw?
(show work)

Accepted Answer

### Problem 8
# Explanation:
## Step1: Recall Torque Formula
Torque $\tau$ is given by $\tau = rF\sin\theta$, where $r$ is the distance from the pivot (forearm length), $F$ is the force, and $\theta$ is the angle between $r$ and $F$. Here, the forearm is at $15^\circ$ below horizontal, and the force to lift is vertical (since lifting is against gravity, force is upward, so the angle between $r$ (forearm, $15^\circ$ below horizontal) and $F$ (vertical) is $90^\circ + 15^\circ= 105^\circ$? Wait, no: if the forearm is at $15^\circ$ below horizontal (so direction of $r$ is $15^\circ$ below x - axis), and the force to lift the dumbbell is upward (vertical, along y - axis). The angle between $r$ (vector from elbow to hand, $15^\circ$ below x - axis) and $F$ (upward y - axis) is $90^\circ+ 15^\circ = 105^\circ$? Wait, no, maybe better: the torque due to a force about a point is $rF\sin\theta$, where $\theta$ is the angle between the position vector $r$ and the force vector $F$. When lifting, the force is vertical (let's assume upward), and the position vector $r$ (forearm) is at $15^\circ$ below horizontal. So the angle between $r$ (direction $15^\circ$ below x - axis) and $F$ (direction positive y - axis) is $90^\circ + 15^\circ=105^\circ$? Wait, no, actually, when you do a bicep curl, the force from the bicep is pulling up, and the forearm is at an angle. Wait, maybe the angle between the forearm (which is the lever arm, length $r = 0.366$ m) and the force (which is vertical) is $90^\circ - 15^\circ=75^\circ$? Wait, let's draw a diagram: horizontal is x - axis, forearm is $15^\circ$ below x - axis (so angle with x - axis is $- 15^\circ$ or $345^\circ$), force is upward (along y - axis, $90^\circ$ from x - axis). The angle between $r$ (at $-15^\circ$) and $F$ (at $90^\circ$) is $90^\circ-(-15^\circ)=105^\circ$? Wait, no, the formula for torque is also equal to $rF\sin\theta$, where $\theta$ is the angle between the lever arm and the line of action of the force. Alternatively, if we consider the perpendicular component of the force to the lever arm. The lever arm is at $15^\circ$ below horizontal, so the perpendicular distance from the elbow to the line of action of the force (if the force is vertical) is $r\sin(90^\circ + 15^\circ)$? No, maybe I made a mistake. Wait, actually, when you lift a dumbbell, the force due to the weight is downward, and the bicep force is upward. But the torque about the elbow: the position vector $r$ is from elbow to hand (length $0.366$ m, at $15^\circ$ below horizontal). The force on the hand due to the dumbbell is downward (weight), but Tamara is applying an upward force of $155$ N to lift it, so the force $F = 155$ N is upward. So the angle between $r$ (direction $15^\circ$ below x - axis) and $F$ (direction positive y - axis) is $90^\circ+15^\circ = 105^\circ$. But $\sin(105^\circ)=\sin(60^\circ + 45^\circ)=\sin60^\circ\cos45^\circ+\cos60^\circ\sin45^\circ=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659$. Alternatively, maybe the angle between the lever arm and the vertical is $15^\circ$, so the angle between $r$ and $F$ is $90^\circ - 15^\circ = 75^\circ$? Wait, no, let's think again. The torque is also equal to the force times the perpendicular distance from the pivot to the line of action of the force. The line of action of the lifting force (upward) is vertical. The perpendicular distance from the elbow (pivot) to this vertical line is $r\cos(15^\circ)$? Wait, no, if the forearm is at $15^\circ$ below horizontal, the horizontal distance from the elbow to the hand is $r\cos(15^\circ)$, and vertical distance is $r\sin(15^\circ)$ below the elbow. The line of action of the upward force is vertical, so the perpendicular distance from the elbow to this line is the horizontal distance, which is $r\cos(15^\circ)$? Wait, I'm confused. Wait, the formula for torque is $\tau=rF\sin\theta$, where $\theta$ is the angle between $r$ and $F$. Let's define $r$ as the vector from elbow to hand: magnitude $r = 0.366$ m, direction $15^\circ$ below x - axis (so angle $\alpha=- 15^\circ$ from x - axis). Force $F$ is upward, so direction $\beta = 90^\circ$ from x - axis. The angle between $r$ and $F$ is $\theta=\beta-\alpha=90^\circ-(-15^\circ)=105^\circ$. So $\tau=rF\sin\theta$.

## Step2: Plug in the values
$r = 0.366$ m, $F = 155$ N, $\theta = 105^\circ$
$\sin(105^\circ)=\sin(60^\circ + 45^\circ)=\sin60^\circ\cos45^\circ+\cos60^\circ\sin45^\circ=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659$
$\tau=0.366\times155\times\sin(105^\circ)$
First, $0.366\times155 = 0.366\times150+0.366\times5=54.9 + 1.83 = 56.73$
Then, $56.73\times0.9659\approx56.73\times0.966\approx54.8$ N·m

Wait, but maybe I got the angle wrong. Alternatively, if the force is vertical (upward) and the forearm is at $15^\circ$ below horizontal, the angle between the forearm (which is the lever arm) and the vertical force is $90^\circ - 15^\circ = 75^\circ$? Wait, no, the lever arm is along the forearm, and the force is vertical. So the angle between the lever arm (forearm) and the force (vertical) is $90^\circ+15^\circ = 105^\circ$ (if forearm is below horizontal) or $90^\circ - 15^\circ=75^\circ$ (if we consider the angle between the lever arm and the perpendicular to the force). Wait, no, the torque formula is $rF\sin\theta$, where $\theta$ is the angle between $r$ and $F$. If the force is vertical and the lever arm is at $15^\circ$ below horizontal, then the angle between $r$ (below horizontal) and $F$ (upward vertical) is $90^\circ + 15^\circ=105^\circ$, as we had before. So the calculation seems right.

# Answer (Problem 8):
Approximately $\boldsymbol{54.8}$ N·m (or more accurately, let's calculate $\sin(105^\circ)$ more precisely: $\sin(105^\circ)=\sin(75^\circ)\approx0.9659$, so $0.366\times155\times0.9659 = 0.366\times155 = 56.73$, $56.73\times0.9659\approx54.8$ N·m)

### Problem 9
# Explanation:
## Step1: Analyze the Forces on the Sled
The combined weight $W = 90$ pounds acts vertically downward. The hill is inclined at $15^\circ$. We can resolve the weight into two components: one parallel to the hill ($F_{\parallel}$) and one perpendicular to the hill ($F_{\perp}$). To keep the sled from sliding down, the force required $F$ must balance the parallel component of the weight.

The formula for the component of the weight parallel to the incline is $F_{\parallel}=W\sin\theta$, where $\theta = 15^\circ$ is the angle of inclination.

## Step2: Calculate the Parallel Component
We know that $W = 90$ pounds and $\theta=15^\circ$.
$F = W\sin\theta=90\times\sin(15^\circ)$
$\sin(15^\circ)=\sin(45^\circ - 30^\circ)=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.2588$
$F = 90\times0.2588\approx23.3$ pounds

# Answer (Problem 9):
Approximately $\boldsymbol{23.3}$ pounds

### Problem 10
# Explanation:
## Step1: Recall Work Formula
The work $W$ done by a force $F$ over a displacement $d$ at an angle $\theta$ to the displacement is given by $W = Fd\cos\theta$.

## Step2: Plug in the Values
Here, $F = 70$ N, $d = 25$ m, $\theta = 50^\circ$
$W=70\times25\times\cos(50^\circ)$
$\cos(50^\circ)\approx0.6428$
$70\times25 = 1750$
$W = 1750\times0.6428\approx1124.9$ joules

# Answer (Problem 10):
Approximately $\boldsymbol{1125}$ joules (or more precisely, $1750\times0.6427876097\approx1124.9\approx1125$ J)

### Problem 11
# Explanation:
## Step1: Resolve the Velocities into Components
The baseball player's velocity: $v_{player}=3$ m/s in the positive x - direction (since running forward, let's assume x - direction is forward).

The ball's velocity: $v_{ball}=20$ m/s at an angle of $20^\circ$ with the horizontal. So the x - component of the ball's velocity is $v_{ball,x}=20\cos(20^\circ)$, and the y - component is $v_{ball,y}=20\sin(20^\circ)$.

## Step2: Find the Resultant x - and y - Components of the Throw
The resultant x - component $v_x$ is the sum of the player's x - velocity and the ball's x - velocity:
$v_x=3 + 20\cos(20^\circ)$
$\cos(20^\circ)\approx0.9397$, so $20\cos(20^\circ)\approx18.794$
$v_x=3 + 18.794 = 21.794$ m/s

The resultant y - component $v_y$ is just the ball's y - velocity (since the player has no vertical velocity):
$v_y=20\sin(20^\circ)$
$\sin(20^\circ)\approx0.3420$, so $v_y = 20\times0.3420 = 6.84$ m/s

## Step3: Find the Resultant Speed
The resultant speed $v$ is given by the magnitude of the resultant velocity vector:
$v=\sqrt{v_x^2 + v_y^2}$
$v=\sqrt{(21.794)^2+(6.84)^2}=\sqrt{475.0 + 46.7856}=\sqrt{521.7856}\approx22.84$ m/s

## Step4: Find the Direction of the Throw
The direction $\theta$ (angle with the horizontal) is given by $\tan\theta=\frac{v_y}{v_x}$
$\tan\theta=\frac{6.84}{21.794}\approx0.3138$
$\theta=\arctan(0.3138)\approx17.4^\circ$

# Answer (Problem 11):
Resultant speed: approximately $\boldsymbol{22.8}$ m/s, direction: approximately $\boldsymbol{17.4^\circ}$ above the horizontal.

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QUESTION IMAGE

Question

Explanation:

Problem 8

Step1: Recall Torque Formula

Step1: Analyze the Forces on the Sled

Step2: Calculate the Parallel Component

Step1: Recall Work Formula

Step2: Plug in the Values

Answer:

Problem 9