What is the weight of the sled being pulled up a ramp by a girl?

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In summary, the conversation discusses a girl pulling a sled up a slippery slope with a given coefficient of friction and maximum acceleration before slipping. The conversation also includes calculations and a question about determining the weight of the sled.
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Statik
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Homework Statement



A girl of mass mg = 50 kg is pulling a sled up a slippery slope. The coefficient of friction between the girl's boots and the slope is µs = 0.155; the friction between the sled and the slope is negligible. The girl can pull the sled up the slope with a ≤ amax = 0.030 m/s2 before she begins to slip. Assume the rope connecting the girl to the sled is kept parallel to the slope at all times. The angle of the slope is θ = 8°.

Homework Equations





The Attempt at a Solution



I have already worked out the maximum value of the force of friction on the girl as:
50kg *9.8 m/s^2 = 490N
Mg(j-hat) = COS 8 * 490N = 485.23N
Fs = Us * Mg(jhat) = 0.155 * 485.23N = 75.21N
Gravity Force on Girl, down the ramp = Sin 8 * 490N = 68.19N

Net force unused by girl is75.21N - 68.19N = 7.02N
Net acceleration is 0.03m/s^2 so, using remaining force, 7.02N / 0.03m/s^2 = 234kg
Total box should be 234kg / sin 8 = 1681 kg but that isn't right. I know my total friction for the girl and I know it's right . . . what am I missing in my calculation of the box weight?

Thanks!
Rod
 
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  • #2
Statik said:

Homework Statement



A girl of mass mg = 50 kg is pulling a sled up a slippery slope. The coefficient of friction between the girl's boots and the slope is µs = 0.155; the friction between the sled and the slope is negligible. The girl can pull the sled up the slope with a ≤ amax = 0.030 m/s2 before she begins to slip. Assume the rope connecting the girl to the sled is kept parallel to the slope at all times. The angle of the slope is θ = 8°.

Homework Equations


The Attempt at a Solution



I have already worked out the maximum value of the force of friction on the girl as:
50kg *9.8 m/s^2 = 490N
Mg(j-hat) = COS 8 * 490N = 485.23N
Fs = Us * Mg(jhat) = 0.155 * 485.23N = 75.21N
Gravity Force on Girl, down the ramp = Sin 8 * 490N = 68.19N

Net force unused by girl is75.21N - 68.19N = 7.02N
Net acceleration is 0.03m/s^2 so, using remaining force, 7.02N / 0.03m/s^2 = 234kg
Total box should be 234kg / sin 8 = 1681 kg but that isn't right. I know my total friction for the girl and I know it's right . . . what am I missing in my calculation of the box weight?

Thanks!
Rod
You did well I in determining the tension in the rope of 7 N, but then you did not analyze the net force acting on the sled correctly. The tension acts up the slope on the sled, but the component of its unknown weight acts down the slope. Try your F_net = ma equation again with a good free body diagram. Like you did when analyzing the forces on the girl.
 

1. How does the angle of the ramp affect the force required to pull the box up?

The angle of the ramp affects the amount of effort required to pull the box up. As the angle of the ramp increases, the force needed to pull the box increases as well. This is because the steeper the ramp, the greater the component of the weight of the box acting against the direction of motion.

2. Does the weight of the box affect the force needed to pull it up the ramp?

Yes, the weight of the box has a direct impact on the force required to pull it up the ramp. The heavier the box, the more force is needed to overcome its weight and pull it up the ramp.

3. Is it easier to pull a box up a longer or shorter ramp?

Generally, it is easier to pull a box up a longer ramp. This is because a longer ramp provides a more gradual incline, which reduces the force needed to pull the box. On the other hand, a shorter ramp with a steeper incline requires more force to overcome the weight of the box.

4. How does friction affect the force needed to pull a box up a ramp?

Friction plays a significant role in determining the force required to pull a box up a ramp. The presence of friction between the box and the ramp increases the force needed to pull the box. This is because friction acts in the opposite direction to the motion, making it harder to move the box up the ramp.

5. Does the type of surface of the ramp affect the force needed to pull the box up?

Yes, the type of surface of the ramp can affect the force needed to pull a box up. A rough or uneven surface can increase the amount of friction between the box and the ramp, making it harder to pull the box. On the other hand, a smooth surface can reduce friction and make it easier to pull the box up the ramp.

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