Physics question help please...?

A late traveler rushes to catch a plane, pulling a suitcase with a force directed 30.0° above the horizontal. If the horizontal component of the force on the suitcase is 60.6 N, what is the force exerted on the handle?





A. 53.0 N


B. 70.0 N


C. 65.2 N


D. 95.6 N














A car goes forward along a level road at constant velocity. The additional force needed to bring the car into equilibrium is





A. greater than the normal force times the coefficient of static friction.


B. equal to the normal force times the coefficient of static friction.


C. the normal force times the coefficient of kinetic friction.


D. zero.

Physics question help please...?
B. 70.0 N


C. the normal force times the coefficient of kinetic friction





The suitcase handle at 30 degrees means that the height of the end off of the ground is half as long as the strap since that ratio is the sine of that 30 degree angle, or 0.5. That gives us a right triangle with legs of (3)^1/2 [square root of three] and one, and a hypotenuse of 2. The hypotenuse of this triangle is 70, with the legs being 35 and 60.6 since (70)^2 = (60.6)^2 + (35)^2





The car must add the power lost to friction, which is the car's weigh times the coefficient of friction while moving (kinetic).
Reply:First Question: I drew it out into a vector diagram with all the forces and their directions, conveniently they came out as a right angled triangle. I then labeled the Horizontal 60.6 and used 60.6*Cos(30) = 52.5 which can be rounded to 53.0. This gives the resultant of the vector and therefore is the force being applied to the handle as it is the resultant.





Second Question: The car is already at equilibrium as it it traveling at a constant velocity, this means that all the forces acting upon it are balanced and therefore in equilibrium.


Hope this helps :)