A firefighter with a weight of 703 N slides down a vertical polewith an accelera

Question

A firefighter with a weight of 703 N slides down a vertical polewith an acceleration of 2.89 m/s2, directed downward. (a) What are the magnitude and direction of thevertical force (use up as the positive direction) exerted by thepole on the firefighter?
1 N

(b) What are the magnitude and direction of the vertical force (useup as the positive direction) exerted by the firefighter on thepole?
2 N (a) What are the magnitude and direction of thevertical force (use up as the positive direction) exerted by thepole on the firefighter?
1 N

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

Explanation / Answer

a) let the force is F (let up be positive), we have: F-mg = ma F=ma+mg = (703N/9.8m/s2)*-2.89m/s2+703N= 495.69N the force is 495.69N, direction is upward. b) magnitude =495.69N, direction downward.

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