When an object is resting on an incline, the down-ramp component of the weight is greater than the frictional force. When an object is resting on an incline, the down-ramp component of the weight is less than the frictional force.
What can you conclude from your observation? When an object is resting on an incline, the down-ramp component of the weight is equal to the frictional force. Refer to the value of the frictional force from the previous parts. F, = u mg cos 0 mg sino o F = 43.8 N Submit Previous Answers Correct Part F - What's the conclusion? < Now that you have calculated the component of the weight of the object that "pushes" itself down the ramp (I call it the down-ramp component of the weight). What is the magnitude of this force for the 11 kg toy car resting on an inclined plane at an angle of 24° with respect to the horizontal? Please express your answer to three significant figures. In order for the object that is resting on the incline to be resting on the incline without sliding down, the frictional force must be equal to that of this force. The magnitude of this force is equal to mg sin 0 as seen in the figure. As an object is resting on a ramp (an inclined plane), there is a component of weight that is causing the object to slide down the inclined plane. Part E - Componet of weight along the direction of the ramp. The frictional force decreases after the coefficient of friction increases. The frictional force on the toy car remains the same even after the coefficient of friction changes. Ff = 43.8 Submit Previous Answers Correct Part D - What can you conclude? After the previous simulations, what can you conclude from what you have done? The frictional force increases after the coefficient of friction increases. Does the frictional force (total) change? < What is the frictional force on the toy car? Please express your answer to three significant figures.
With the slope of the inclined plane remains the same, change the coefficient of frictions of the car and of the ground to 0.8. Use the same Algodoo scene you have just created. If the car moves, the kinetic frictional force will be different if the coefficient of friction is different. Submit Previous Answers Correct Please notice that this conclusion only works if the car does not move because the frictional force will still be equal to the component of the weight of the car parallel to the inclined plane, mg sin 0, which is the same regardless of what the coefficient of friction is. What do you think the frictional force on the tires will become? Becomes higher. ► View Available Hint(s) F = 43.8 N Submit Previous Answers Correct Part B - What do you think? Suppose now that the coefficient of friction becomes 0.8 and the car still does not move. Use the values you find in the Algodoo scene to find the total frictional force on the toy car? Please express your answer in three significant figures.
Those are the frictional forces on the tires. As soon as you click the "play" button, you will see two arrows labeled "T". Please double check to make sure that the coefficient of friction of the car and the ground are set to 0.7 or else your simulation will be wrong. Change the mass of the car to 11 kg, the coefficients of friction of the car and the ground to 0.7, and use the rotation tool to change the angle to match what is given above.
#Algodoo car download#
Please download the Algodoo scene from Canvas. To simply your effort, an Algodoo scene has been created. To simulate this situation we can create an Algodoo scene and observe the forces in that world. We will create a toy car in the Algodoo world and simulate this. What is the total frictional force on the car? We will answer this question by way of Algodoo. Please note that in the Algodoo world, the coefficient of static and kinetic friction are the same. The coefficient of friction between the tires and the road is 0.7. Transcribed image text: v Part A A 11 kg toy car is parked on a 24° ramp.
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