Question

6. A 600 N person sits on a 50 N sled. The coefficient of static friction is 0.14. How much force is needed, respectively, to start the motion by pushing [A] and pulling [B] at an angle of 30°? (Rounding to the nearest integer)

Answer 1

A.)

By pushing,

By force balance in vertical direction,

Normal force(N) = F*sinA + W

By force balance in horizontal direction,

F*cosA = Fs

here, F = minimum pushing force = ??

A = angle = 30 deg

Fs = friction force = s*N

s = static friction co-efficient = 0.14

W = total weight = weight of person + weight of sled = 600 N + 50 N = 650 N

So, from above equations,

F*cosA = s*(F*sinA + W)

F*cos 30 deg = F*0.14*sin(30 deg) + 0.14*650

F = 0.14*650/(cos(30 deg) - 0.14*sin(30 deg))

F = 114.32 N

Rounding off to nearest integer

F = 114 N

B.)

By pulling,

By force balance in vertical direction,

Normal force(N) = -F*sinA + W

By force balance in horizontal direction,

F*cosA = Fs

here, F = minimum pulling force = ??

So, from above equations,

F*cosA = s*(-F*sinA + W)

F*cos 30 deg = -F*0.14*sin(30 deg) + 0.14*650

F = 0.14*650/(cos(30 deg) + 0.14*sin(30 deg))

F = 97.22 N

Rounding off to nearest integer

F = 97 N

Let me know if you've any query.