Newton's law and friction

[simor]

In the diagram below, the cord makes a 25° angle with the horizontal, the mass of the sled and occupants is 100 kg. The tension in the cord is 120 N and the friction force is 15 N. Find the acceleration of the sled.

So far I've found the components of the cord tension by drawing a triangle and using trig:
Tx = T cos(25°) = 108.757
Ty = T sin(25°) = 50.7142

To find weight, I used mg:
W = m g = 100*9.8 = 9800

Then I used the horizontal force equation to find acceleration
Tx - fk = m a
108.757 - 15 = 100 * a

So I ended up with acceleration of 109, which I feel is too large. Is that correct? I don't have an answer key so I'm not sure.

[jeff]

Hi simor,

In the diagram below, the cord makes a 25° angle with the horizontal, the mass of the sled and occupants is 100 kg. The tension in the cord is 120 N and the friction force is 15 N. Find the acceleration of the sled.

So far I've found the components of the cord tension by drawing a triangle and using trig:
Tx = T cos(25°) = 108.757
Ty = T sin(25°) = 50.7142

To find weight, I used mg:
W = m g = 100*9.8 = 9800

Then I used the horizontal force equation to find acceleration
Tx - fk = m a
108.757 - 15 = 100 * a

So I ended up with acceleration of 109, which I feel is too large. Is that correct? I don't have an answer key so I'm not sure.

No, that's not quite right. We can see that the acceleration has to be less than 1.08, because that would be the acceleration if there were no friction.

However, your last equation that you show is correct, this one:

108.757 - 15 = 100 * a

so I think you might have just entered it into the calculator incorrectly. Did you make sure to subtract the numbers on the left first, and then divide that result by 100? After subtracting, the next step would be:

93.7569 = 100 * a

[simor]

You're right, I entered it incorrectly in my calculator. I get 0.938 now.