Answer:
[tex]F_H_n=230.04 N[/tex]
The Required horizontal force is 230.04N
Explanation:
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
[tex]F_H_n=F_f\\F_h=mg*u[/tex]
where:
F_{Hn} is the required Force
u is the friction coefficient
m is the mass
g is gravitational acceleration=9.8m/s^2
[tex]200=mg*u[/tex] Eq (1)
Now, weight increases by 42% and friction coefficient decreases by 19%
New weight=(1.42*m*g) and new friction coefficient=0.81u
[tex]F_H=(1.42m*g*.81u)[/tex] Eq (2)
Divide Eq(2) and Eq (1)
[tex]\frac{F_H_n}{200}=\frac{1.42m*g*0.81u}{m*g*u}\\F_H_n=1.42*0.81*200\\F_H_n=230.04 N[/tex]
The Required horizontal force is 230.04N