Question

6) The 55 kg cart below is pulled with a Force, P, towards the left at the angle shown. Answer the following questions a) Draw a free body diagram and a kinetic diagram of the cart. (3 points) D.2 m HA let 0.15 m 0.25 m b) If the total moment caused by Force P is in the clockwise direction, what is the largest force that can be applied without lifting the wheels located at Point A off the ground? (3 points) c) What is the corresponding normal force at Point B? (3 points) d) What is the corresponding acceleration of the cart? (3 points)

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Answer 1

Solution: - e body diagram: Pe apy a weight w=55X9a81 fig:FBD ६ joke o o diagram - W = 539.55N Poc Px = p c0535 = 0·82P Py=psin35=0.57P NA and Npare normal force at wheel. acceleration was shows in diagram for kinetic diagram. Equation of motion- fig: kinetic Px = ma → la-Ryon > Acceleration &a", o b) : Net moment caused by force p will be in clockwise direction, hence point A will lift first, hence to calculate largest force, NATO (Just before lifting) Now we will balance the net moment caused by all forces about point B=0;- .57P 620-25m, c= 0.8m, d=0.50 Now EMB=01- a 0.82p t => wxb-0.57P8C to: 82P xdzo >> 539:55X0-25-0-57PX0-8+0.82PX0-5-0 → [P= 155.76N G largest force di --- NAFO No

CULIS 9 for Normal force at B- Efy=01 NB-W to 570=0 >> Ne = 539.55-0.57X155.76 »[NB= 450.77 N Ang d) Acceleration of cast - Using Eanu a= man = 0:82X 155.76 a= 2.322 m/s?) Ang
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