1) The Force Table (revisited)- I am giving some of you another shot at Case 3 o
ID: 3281159 • Letter: 1
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1) The Force Table (revisited)- I am giving some of you another shot at Case 3 only this time it's not a "nice" triangle. If that somehow complicated things for you last time, this should be a breeze then! The given values are as follows: T, = 726N, Tb=453N, Tc-631N and ,-12.18 Find ,and .using the clearly suggested technique from the lab manual. Show all work, unclear solutions will not be considered. Give the solution angles with 6 significant figures. 2) Unnecessary details from Work Energy- Derive for me the acceleration (a) of the glider for the setup used in part 2 of the Work Energy lab (on an incline) as a function of only (or at most as some may not be necessary) the following variables: the glider mass (M), the hanging mass (m), the angle of inclination (e), the velocity of the glider (v), the displacement of the glider (s), and the gravitational acceleration (g). Show all free body diagrams and steps. 3) Friction rubbing the wrong way?-Determine the static friction coefficient based on part 2 of the friction lab given the following conditions and outcome: The angle of inclination is exactly 12°, the mass of the kart (with mass inside included) is measured to be 1,216.000g and the value for the maximum tensiorn force in my string as measured by my force sensor is 9.704N and the value of g is exactly 9.8.Give your answer with 3 significant figures.Explanation / Answer
1. given
Ta = 726 N
Tb = 453 N
Tc = 631 N
thetaa = 12.18 deg
so let the other two angles be thetab and thetac
in vector notation the forces can be written as
Ta = 726(cos(12.18)i + sin(12.18)j) [ where i and j are unit vectors in x and y directions respsectively]
Tb = 453(cos(thetab)i + sin(thetab)j)
Tc = 631(cos(thetac)i + sin(thetac)j)
for equilibrium
Ta + Tb + Tc = 0
726(cos(12.18)i + sin(12.18)j) + 453(cos(thetab)i + sin(thetab)j) + 631(cos(thetac)i + sin(thetac)j) = 0
comparing coefficients
453cos(thetab) = - 631cos(thetac) -709.6574
453sin(thetab) = - 631sin(thetac) -153.174
squaring and adding
205209 = 398161 + 527075.89965076 + 895587.6388cos(thetac) + 193305.588*sin(thetac)
-720027.89965076 - 895587.6388cos(thetac) = 193305.588*sin(thetac)
squaring both sides
518440176275.4849125299685776 + 802077218771.35926544*cos^2(thetac) + 1289696173036.694986050976cos(thetac) = 37367050352.025744 - 37367050352.025744cos^2(thetac)
839444269123.38500944cos^2(thetac) + 1289696173036.694986050976cos(thetac) + 481073125923.4591685299685776 =0
soving for thetac
thetac = 129.621 deg
thetab = 132.709 deg
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