0/10 points | Previous Answers OSColPhys1 4.P.029.Tutorial.WA.My Notes | Questio

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0/10 points | Previous Answers OSColPhys1 4.P.029.Tutorial.WA.My Notes | Question Part Points Submissions Used A contestant in a winter games event pushes a 36.0-kg rock across a frozen lake with a force of 25 N at 26? below the horizontal as shown in Figure (a) below, and it moves with an acceleration of 0.62 m/s2 to the right. (a) What is the normal force exerted by the lake surface on the rock? Did you draw a free-body diagram and identify the forces acting on the system? Did you consider all applicable forces (or component of forces) in the vertical direction? N (b) Instead of pushing on the rock, the contestant now pulls on it with a rope over his shoulder at the same angle above the horizontal as in part (a). See Figure (b) above. Now what is the normal force exerted by the lake surface on the rock? N

Explanation / Answer

Normal force will act in vertical direction which will balance gravity
a) Now if lets assume boy applies a force of F newton on block then,

a) vertical component of force will be downwards
thus, and horzontal component will give acceleration
Fcos = ma

=> 26 cos = 36*0.62

=> = 30.855 degrees

Thus in vertical direction

Fsin + mg = N (as there isnoacceleration invertical)

Normal force = 26 sin30.855 + 36*9.8 =366.135 N

b) Now in this case as force is pullingnot pussing thus its vertical component will be upwards

but there is no other force in horizontal hence,

to produce same amount of acceleration he will have to pull rope at the same angle as in first case

But equation in vertical becomes

N=mg - Fsin30.855 = 36*9.8-26*sin30.855   = 339.465 N

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