David C. answered • 09/02/20
Retired Physics Teacher (28 years)
First of all, THERE IS NOT ENOUGH INFORMATION TO ANSWER ALL THE QUESTIONS (UNLESS YOU ASSUME THE CART IS MOVING AT A CONSTANT VELOCITY (a=0m/s2))
You will need to break the original applied force vector into its components:
280N applied downward and forward at a 15 degree angle would result in a forward and downward component.
MAKE SURE YOUR CALCULATOR IS IN DEGREES MODE.
Forward can be found by drawing a triangle with an hypotenuse of 280N and an angle of 15 degrees. The horizontal leg of the triangle would be found using Cosine. Cos 15 = ADJ / 15 >>> ADJ = 15 * Cos 15 = 14.49N. (14N forward in significant figures)
The downward force is found by finding the opposite side of the triangle. Sin 15 = OPP / 15 >>> OPP = 15 * Sin 15 = 3.88N. (3.9N downward in significant figures)
The objects weight is simply how hard gravity will pull on it. Using 9.81 m/s2 for a (g) and using F=ma or F=mg >> 25kg * 9.81m/s2 = 245N. (250N in significant figures)
The normal force of the cart is the total downward force into the floor (perpendicular). All you need to do is add up all downward forces. 245N from gravity and 3.9 from the person's downward component of their push = 245N + 3.9N = 248.9N (again, it's 250N in significant figures).
Now, if they would have said the cart was moving at a constant velocity, aka not accelerating, we could answer the rest of the questions. If a=0m/s2, then the net force on the cart is 0N...F=ma . F=25 * 0 = 0. That means the total friction of the cart must be equal to the total forward push, thus canceling each other. So the total friction of the cart is 14N in significant figures.
