A firefighter with a weight of 702 N slides down a vertical pole with anaccelera

Question

A firefighter with a weight of 702 N slides down a vertical pole with anacceleration of 2.97 m/s2, directed downward. (a) What are the magnitude and direction of the vertical force (useup as the positive direction) exerted by the pole on thefirefighter?
______ N

(b) What are the magnitude and direction of the vertical force (useup as the positive direction) exerted by the firefighter on thepole?
_______ N (a) What are the magnitude and direction of the vertical force (useup as the positive direction) exerted by the pole on thefirefighter?
______ N

(b) What are the magnitude and direction of the vertical force (useup as the positive direction) exerted by the firefighter on thepole?
_______ N

Explanation / Answer

a) mass of man is m=702/9.8 =71.6 We have friction in the system. Using newtons laws ofmotion: mg -Friction force = ma Friction =702- 71.6*2.97 =489.3 N This is the force applied by pole to the firefigther
b) this is simply weigth of person = 702 N

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