Final answer:
The static friction force will be 75.35 N when the external force is half of the minimal force required to overcome the static friction, assuming the static friction force can reach this value before the object starts moving.
Step-by-step explanation:
When the external force applied is half the minimal force necessary to overcome static friction, which was previously calculated at 150.7 N, the new force applied is f = fmin/2 = 150.7 N / 2 = 75.35 N. Static friction is a self-adjusting force that matches the external force applied up to its maximum value. The maximum static frictional force is given by fs(max) = μs N, where μs is the coefficient of static friction and N is the normal force. From the reference information provided, we know the coefficient of static friction is μs = 0.377, and the normal force N = wp = 400.0 N. Therefore, the maximum possible static frictional force is fs(max) = 0.377 * 400.0 N = 150.7 N. Since the applied force of 75.35 N is less than the maximum static frictional force, the static friction force in this case will be 75.35 N, equal to the applied force, but in the opposite direction, keeping the object stationary.