A load is pulled at a uniform speed along a horizontal floor by a rope at 45^o to floor. If the force in the rope is 1500N, what is the frictional force on the load ?

Okonkwo_Uche

4 months ago

how are we meant to do this in jamb if there are no trig functions?

odewenusofiyyah

2 years ago

Ht

Sammyporsche123

2 years ago

I don't understand

igwomali

2 months ago

I don't understand this what's this pls

sonialyzo1

1 year ago

To find the frictional force on the load, let's go through the problem step-by-step.

### **Given:**
- Force in the rope, \( F = 1500 \text{ N} \)
- Angle with the floor, \( \theta = 45^\circ \)
- Load is moving at a **uniform speed** (implies no net acceleration, so the frictional force must balance the horizontal component of the pulling force).

### **Step 1: Find the Horizontal Component of the Force**
The frictional force must balance the horizontal component of the force since the speed is uniform.

\[
F_{\text{horizontal}} = F \cos \theta = 1500 \cos 45^\circ
\]

Since \( \cos 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707 \), we get:

\[
F_{\text{horizontal}} = 1500 \times 0.707 = 1060.5 \approx 1061 \text{ N}
\]

### **Conclusion:**
Since the load is moving at a uniform speed, the **frictional force** must equal the horizontal component of the applied force.

**Correct Answer: D. 1061N**

Sammyporsche123

2 years ago

cos 45 is 0.760