A tired and oveloaded is attempting to hold a large physics textbook wedged unde

Question

A tired and oveloaded is attempting to hold a large physics textbook wedged under his arm, as shown above.

The textbook has a mass of 3.1 kg, while the coefficient of static friction of the textbook against the student's underarm is 0.320 and the coefficient of static friction of the book against the student's shirt is 0.160.
(a) What is the minimum horizontal force that the student must apply to the textbook to prevent it from falling?
1 N

(b) If the student can only exert a force of 62.6 N, what is the acceleration of the textbook as it slides from under his arm? The coefficient of kinetic friction of arm against textbook is 0.200, while that of shirt against textbook is 0.090. magnitude 2 m/s2
direction 3 ---Select--- upward/downward

A tired and oveloaded is attempting to hold a large physics textbook wedged under his arm, as shown above. The textbook has a mass of 3.1 kg, while the coefficient of static friction of the textbook against the student's underarm is 0.320 and the coefficient of static friction of the book against the student's shirt is 0.160. (a) What is the minimum horizontal force that the student must apply to the textbook to prevent it from falling? 1 N (b) If the student can only exert a force of 62.6 N, what is the acceleration of the textbook as it slides from under his arm? The coefficient of kinetic friction of arm against textbook is 0.200, while that of shirt against textbook is 0.090. magnitude 2 m/s2 direction 3 ---Select--- upward/downward

Explanation / Answer

a) we can add the coefficients of static friction together, because there is only one force applied. so .480. Now the force of gravity on the book is mg, 3.1 x 9.8 = 30.38 N The force due to friction must be greater or equal to this 30.38 N force. so F(friction) = 30.38 = F x coef of friction F = 30.38 / .480 = 63.29 N b) Now he can only exert a force of 62.6 N, and the coef of friction is .2 + .09 = .29, so the force upward due to friction is 62.6 x .29 = 18.154 N The force on the book downward is 30.38 N Net Force = 30.38 - 18.154 = 12.226 N downward F=ma 12.226 = (3.1)a a = 3.94 m/s^2 Since the force is downward, then the acceleration is downward too a= 3.94 m/s^2 Downward direction

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