A passenger is pulling on the strap of a 15.0-kg suitcase with a force of 60.0 N. The strap makes an angle of 32.5° above the horizontal. A 37.8-N friction force opposes the motion (horizontal) of the suitcase. Determine the acceleration of the suitcase.

Question

anonymous · Answer

Since the force applied to the suitcase is at angle with the horizontal, you have to find the horizontal and vertical components of the force. Since the suitcase only moves horizontally, gravity, normal force, and the vertical component of the applied force can be disregarded. Draw a right triangle with a hypotenuse of 60 N and an angle of 32.5 degrees with the horizontal. Using trigonometry, you can find the horizontal leg of the triangle to be 50.6 N. This is the magnitude of the applied force opposing friction. Find the difference of the two forces (50.6 - 37.8) = 12.8 N.  Since F = ma, a = F/m. Using this equation, divide the resultant force on the suitcase (12.8 N) by its mass (15 kg) to find an acceleration of a = 0.853 m/s^2.