Alright, let’s go through these questions step by step.
### Part A:
To find the magnitude of the torque when the force is exerted perpendicular to the door, we use the following formula:
[tex]\[ \text{Torque} (\tau) = \text{Force} (F) \times \text{Lever arm} (r) \][/tex]
Given:
- Force, [tex]\( F = 40 \)[/tex] N
- Width of the door, [tex]\( r = 92 \)[/tex] cm, which we need to convert to meters. Hence, [tex]\( r = 92 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 0.92 \, \text{m} \)[/tex]
Now, substituting the given values into the torque formula:
[tex]\[ \tau = 40 \, \text{N} \times 0.92 \, \text{m} = 36.8 \, \text{N} \cdot \text{m} \][/tex]
To express the answer using two significant figures:
[tex]\[ \tau \approx 37 \, \text{N} \cdot \text{m} \][/tex]
So, the magnitude of the torque when the force is exerted perpendicular to the door is:
[tex]\[ \boxed{37 \, \text{N} \cdot \text{m}} \][/tex]
### Part B:
To find the magnitude of the torque when the force is exerted at a 65.0° angle to the face of the door, we use the formula that incorporates the angle:
[tex]\[ \text{Torque} (\tau_{\theta}) = \text{Force} (F) \times \text{Lever arm} (r) \times \cos(\theta) \][/tex]
Given:
- Force, [tex]\( F = 40 \)[/tex] N
- Width of the door, [tex]\( r = 0.92 \, \text{m} \)[/tex]
- Angle, [tex]\( \theta = 65° \)[/tex]
- We first convert the angle from degrees to radians to use it in the cosine function.
Converting [tex]\( \theta \)[/tex] to radians:
[tex]\[ \theta_{\text{rad}} = 65 \times \frac{\pi}{180} \approx 1.1345 \text{ radians} \][/tex]
Then, using the torque-angle formula:
[tex]\[ \tau_{\theta} = 40 \, \text{N} \times 0.92 \, \text{m} \times \cos(1.1345) \][/tex]
Given the result we earlier found from our calculations:
[tex]\[ \tau_{\theta} \approx 15.55 \, \text{N} \cdot \text{m} \][/tex]
So, the magnitude of the torque when the force is exerted at a 65.0° angle to the face of the door is:
[tex]\[ \boxed{15.55 \, \text{N} \cdot \text{m}} \][/tex]
