Question

Using a forked rod, a 0.35-kg smooth peg P is forced to move along the vertical slotted path T' = (0.5 0) m, where 0 is in radians. The angular position of the arm is 0 (t)rad where t is in seconds. The peg is in contact with only one edge of the rod and slot at any instant. (Figure 1) Part A Determine the magnitude of the force of the rod on the peg at the instant t = 2 s. Express your answer to three significant figures and include the appropriate units. HA ? F = Value Units Submit Request Answer Figure Part B 1 of 1 Determine the magnitude of the normal force of the slot on the peg. Express your answer to three significant figures and include the appropriate units. PA ? N = Value Units r = (0.50) m Submit Request Answer Provide Feedback

Answer 1

Giren that r=6.50) mass of peg (M)= 0.35 kg vertical slotted path (weight should consider) r=(0.5o)m where o in in rad 2 O = where t is seconds is a)= t = - 25 y=5750 fort=28 = T (2)² o 8 π rad 0909 2 11 909 gia dr 11 (0:50) =d60.50).do dt dt do dt 0.5 (0:50) a) Ir= 0.5

2 O - (= 16 do d dt (It) - T dt (pt) 8 Cot) ò It 4 ö d - d dt de ㅍ t) 4 T 1 - 1 At t=28 o = I rad TP t = radls The T (2) - ។ ö I rad/a² 4

0.5 r= 0.50 = 0.5()=0-25T)M řa = (0.257) "/ ř.0.5 Ö= 0.5 (0.1255 T (2) 0.5o = T) M/62 tan v = 0.50 d. 0.50 do 0= 11 rad do = 0,5 05 8=11 rad tan y = I rad 2 anticlock wire from p=575180 Positive r-axis the system Free body diagram of par p=57.5189 N Tangent 90-57.518 = 32.482° 0 * м9 The angle between tangent 909 and Normal force is Alsume g= 9.81 m/s?

motion Apply equation of Mar +łe Fr + I { Fo= mag raro² ar = 6-1257) - (0.25W) (2) mla? ar - 1.545 m/s2 ao rot zro ॥ (0.25") () + 2 (0-257) ( a 3.0842 m/s2 From FBD tľ Efr= Mar Mar -NCB 32.482 - ng -N COM32.48- 0.35) (9.41) = (6:35) 61.545) N=-3.429 N

+ & {fo= Mag Ft N sin 32.482 = Mao F + (-3.429) Sin32.482 = 6:35) (3.0842) F = 2.9209 N the peg The Magnitude of the force on the instant t=28 CE) = 2.9209N the at the slot The magnitude of Normal force of (A) = -3.429 N on peg