Yusef pushes a chair of mass m = 55.0 k g across a carpeted floor with a force F

Question

Yusef pushes a chair of mass m = 55.0kg  across a carpeted floor with a force F? p (the subscript 'p' here is lowercase and throughout the question) of magnitude Fp = 148N  directed at ? = 35.0degrees below the horizontal (Figure 1) . The magnitude of the kinetic frictional force between the carpet and the chair is Fk = 106N .

What is the magnitude of the acceleration a of the chair? What is the magnitude of the normal force FN acting on the chair?

i found ay which is 0m/s2 and Fg is 539N but i need to find the forces in the y direction to find the trun FN

Explanation / Answer

a = F*cos35- f /m = (148*cos35-106)/55 = 0.28 m/s^2

N = m*g + Fsin35 = (55*9.8)+(148*sin35) = 623.88 N

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.