The ACE towing company tows a disabled 10,500 N automobile off the road at a constant speed. If the tow line makes an angle of 10.0° with the vertical as shown, what is the tension in the line supporting the car?
10,448 N
Well, towing a disabled car sounds like quite a drag! But let's tackle this problem together, shall we? We can use some trigonometry to figure out the tension in the tow line.
Since the line makes an angle of 10.0° with the vertical, we can break down the tension into two components: one in the vertical direction and one in the horizontal direction.
The vertical component of the tension counteracts the weight of the car, which is 10,500 N. So, we get:
Tension_vertical = weight × cos(angle)
Tension_vertical = 10,500 N × cos(10.0°)
Now, let's calculate that:
Tension_vertical = 10,500 N × 0.9848 (approximately)
Tension_vertical = 10,293.6 N (approximately)
So, the tension in the vertical direction is approximately 10,293.6 N.
Now, for the horizontal component, it doesn't really affect the tension in the vertical direction. So, we can ignore it for now and focus on the vertical one.
Therefore, the tension in the line supporting the car is approximately 10,293.6 N.
Hope that helps lighten the load of this problem a bit!
To find the tension in the line supporting the car, we can use trigonometry.
The weight of the car is 10,500 N, which acts vertically downwards. The tow line makes an angle of 10.0° with the vertical, so we need to find the vertical component of the tension force.
The vertical component of the tension force can be found using the equation:
Vertical component of tension force = Tension force * cos(angle)
In this case, the angle is 10.0°, so we have:
Vertical component of tension force = Tension force * cos(10.0°)
To find the tension force, we can rearrange the equation:
Tension force = Vertical component of tension force / cos(angle)
Now let's calculate the tension force:
Tension force = 10,500 N / cos(10.0°)
Using a calculator, we can find the value of cos(10.0°) to be approximately 0.9848.
Tension force = 10,500 N / 0.9848
Tension force ≈ 10,659.20 N
Therefore, the tension in the line supporting the car is approximately 10,659.20 N.
To find the tension in the line supporting the car, we can apply Newton's second law of motion. First, let's resolve the forces acting on the car in the vertical direction.
Since the car is being towed at a constant speed, the sum of the forces in the vertical direction must be zero:
ΣFy = 0
The only vertical force acting on the car is the tension in the tow line, which supports the weight of the car. The weight of an object can be calculated using the formula:
Weight = mass × gravitational acceleration
Since the weight acts downwards, we have:
Weight = -m × g
where m is the mass of the car and g is the gravitational acceleration (approximately 9.8 m/s²).
Now, let's resolve the forces acting on the car in the horizontal direction.
There is no acceleration in the horizontal direction, so the sum of the forces in the horizontal direction must also be zero:
ΣFx = 0
The only horizontal force acting on the car is the horizontal component of the tension force in the tow line.
Now, we can use trigonometry to find the horizontal component of the tension force. We have a right triangle formed by the tow line, the vertical direction, and the angle between them. The horizontal component of the tension force can be found using the equation:
Tension horizontal = Tension × cos(angle)
where Tension is the tension in the tow line and angle is the angle between the tow line and the vertical direction.
Since the horizontal component of the tension force is the only horizontal force acting on the car, it must balance the horizontal component of the gravitational force. Therefore, we have:
Tension horizontal = Weight horizontal
Weight horizontal = Weight × sin(angle)
Finally, we can substitute the values into the equations to calculate the tension in the line supporting the car.
Weight = -m × g
Weight horizontal = Weight × sin(angle)
Tension horizontal = Weight horizontal
Tension = Tension horizontal / cos(angle)
T•cosα=m•g
T=mg/cosα =
=10500/cos10°= 10660 N