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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