Tension in Rope: Truck Towing 1T Car Up Hill

AI Thread Summary
A truck is towing a 1,000 kg car uphill at a constant speed on a 5-degree incline, with a rope attached at a 10-degree angle. The discussion focuses on determining the tension in the rope while neglecting friction. Participants clarify that constant speed implies zero acceleration, meaning the sum of the forces in both the x and y directions must equal zero. The normal force is considered to be equal to the weight of the car, but the challenge lies in resolving the forces along the slope due to their different angles. Ultimately, the tension in the rope can be calculated by analyzing the components of the forces involved.
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A truck is towing a 1.00*10^3 kg car at a constant speed up a hill that makes an angle of 5degrees with respect to the horizontal. A rope (negligable) is attached to the truck at an angle of 10 degrees with respect to the horizontal. Neglect friction. What is the tension in the rope?
I'm not sure if the sum of the x and y components should be zero or not?
 
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milo7979 said:
A truck is towing a 1.00*10^3 kg car at a constant speed up a hill that makes an angle of 5degrees with respect to the horizontal. A rope (negligable) is attached to the truck at an angle of 10 degrees with respect to the horizontal. Neglect friction. What is the tension in the rope?
I'm not sure if the sum of the x and y components should be zero or not?

Hi milo7979! Welcome to PF! :smile:

Constant speed means zero acceleration …

so, yes, the sum of the components in any direction must be zero. :smile:

(though you might find it easier to use components along the slope :wink:)
 
i thought about using different coordinates, but only one of the forces will be on the axis because there are three forces with different angles with respect to the horizontal. I can't seem to get all of the y components to equal zero. The y component of the normal force I assume is the same magnitude as the weight (1.0*10^3 * g= 9.8*10^3N), but there is still a y component to the Force from the truck. W and y component of N equal zero, so what am I missing?
 
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milo7979 said:
i thought about using different coordinated, but only one of the forces will be on the axis because there are three forces with different angles with respect to the horizontal.

Yes, there are three forces, and two of them are along the slope and perpendicular to it … :wink:
 
I don't think that two of the forces are along the x and y axis. N is perpendicular to the hill which is 5 degrees to the left of vertical, W which is straight down the vertical and F of truck pulling the car is 10 degrees above horizontal.
 
milo7979 said:
I don't think that two of the forces are along the x and y axis. N is perpendicular to the hill which is 5 degrees to the left of vertical, W which is straight down the vertical and F of truck pulling the car is 10 degrees above horizontal.

oh, sorry :redface:

I misread it …

ok, the force you want is the tension in the rope, so calulate the normal force, and then take components of everything along the line of the rope. :smile:
 
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