Static Friction - Cart on a horizontal track. Cart on top of horizontal surface
ID: 1328420 • Letter: S
Question
Static Friction - Cart on a horizontal track.
Cart on top of horizontal surface = 650 g
Hanging mass, initial = 50 g
Hanging mass, final = 120 g
Initially, the cart doesn’t move – it’s in equilibrium.
There are four vectors shown acting on it. Horizontally, the tension (T), pulls the cart to the right and the static force of friction (fs ) pulls it with equal force to the left. Vertically, there are two forces: the gravity force which equals to the weight of the cart (Wc = Mg) is going downwards and the normal force (N ) opposed to it. The hanger also has a pair of forces acting on it. The upward tension force (T) is balanced by an equal, downward gravity force which equals to the weight of the hanging mass (Wh). Now start sequentially adding small masses (in steps of 10 grams) to the hanger until the system starts moving. 70 grams were added until the mass began to move, so now the total hanging mass is 120 g.
At the moment of movement tension force equals to the maximum static friction force between the cart and the track T = fs, max = µs N . The tension in the hanger equals to its weight T = Wh= mh g. Combining these two equations, one can solve for µs. The calculated value of µs needs to be compared with the given coefficient of static friction (µs = 0.18) by calculating the discrepancy.
1. Calculate the coefficient of the static friction µs:
2. What is the discrepency between the experimentally determined µs and it's actual given value (µs = 0.18)
Explanation / Answer
1.
T = mh g = 0.120 x 9.8 = 1.176 N
for the cart , Force equation in vertical direction is given as
Fn = Wc = Mg = 0.65 x 9.8 = 6.37 N
Along the horizontal direction:
fs = T
us Fn = T
us (6.37) = 1.176
us = 0.19
b)
discrepency = 0.19 - 0.18 = 0.01
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.