Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
Answer:
A.) 1372 N
B.) 1316 N
C.) 1428 N
Explanation:
Given that a 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following cases:
a. The load moves downward at a constant velocity
At constant velocity, acceleration = 0
T - mg = ma
T - mg = 0
T = mg
T = 140 × 9.8
T = 1372N
b. The load accelerates downward at a rate 0.4 m/s??
Mg - T = ma
140 × 9.8 - T = 140 × 0.4
1372 - T = 56
-T = 56 - 1372
- T = - 1316
T = 1316N
C. The load accelerates upward at a rate 0.4 m/s??
T - mg = ma
T - 140 × 9.8 = 140 × 0.4
T - 1372 = 56
T = 56 + 1372
T = 1428N
Answer:
v = 2000m/60s; v = 33.33 m/s (plus, in velocity direction is needed in answer, if given)
Explanation:
velocity = distance/time
