High School

A car traveling down a road is navigating a sharp turn. Two forces acting on a tire with respect to a certain axis are 82.9 N at an angle of 30 degrees and 97.7 N at an angle of -53 degrees. If these are the only external forces acting on the tire, what is the magnitude and direction of the third friction force, balancing them and preventing the tire from sliding?

For your solution to the question, complete the following steps:

a) Draw a qualitative picture of this situation. Choose an x-axis and sketch in the two known forces. (It is not necessary to make an accurate diagram using a protractor and ruler.)

b) Based on what you have drawn, predict the magnitude and direction of the third vector and add it to the sketch. (It does not have to match the final solution exactly; this is just a sketch.)

c) In a few sentences, describe your problem-solving strategy. What are you going to do, and in what order, to find a solution to this word problem?

d) Solve to find the magnitude and direction of the third force. For each step, begin by writing down either one of the equations we have given you in the header or one that is a result of the problem setup. Next, replace variables in the equation with numbers from the problem. After one or more algebraic steps, record your answer to that step, including a unit on the value and a reasonable number of significant digits (not the full calculator display!).

e) Write down your final answer, both magnitude and direction, to the closest tenth of a Newton and the closest degree.

f) Compare your solution to your prediction. Comment on whether the calculated result appears to match your prediction, and if there are any discrepancies between the two.

Answer :

Final answer:

The magnitude of the third friction force is 133.2 N and it acts at an angle of -16 degrees relative to the x-axis. This force balances the other two, preventing the tire from sliding.

Explanation:

To start, we can represent the forces as vectors in the xy-plane, where positive x goes right and positive y goes up. The force of 82.9 N at an angle of 30 degrees can be represented as (82.9*cos(30), 82.9*sin(30)) = (71.7, 41.5). The force of 97.7 N at an angle of -53 degrees can be represented as (97.7*cos(-53), 97.7*sin(-53)) = (58.5, -78.3).

We understand that the forces balance each other, hence the net force acting on the object is zero. Therefore, the third force (which is implicitly the friction force) must be such that when it's added to these two forces, the result is (0,0). So, we get the third force as F3 = -(F1 + F2) = -((71.7, 41.5) + (58.5, -78.3)) = (-130.2, 36.8) N.

The magnitude of this force can be calculated by sqrt(x^2+y^2) = sqrt((-130.2)^2 + (36.8)^2) = 133.2 N. The direction can be found using atan(y/x) = atan(36.8/-130.2) = -16 degrees (the negative angle meaning it's below the x-axis).

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